Collection of Solved Problems in Physics

Motion on two inclined planes

Task number: 1149

• Hint 1 - to the part A

  In the first part of the task we neglect the moment of inertia of balls – in such a case, is it possible to replace the balls with simpler physical concept?

• Solution Hint 1

  If we neglect the moment of inertia of balls, it is possible to replace balls with point masses.

• Hint 2 – to the part A

  Draw a free body diagram and choose an appropriate reference frame. What does it mean „determine the motion of the centre of mass“? What should be the output of the task?

• Solution Hint 2

  The figure bellow shows the appropriate reference frame used in this problem.

  

  We consider that the motion is situated in the plane xy, thus we can omit the change of z-coordinate. The output of the task should be a formula which describes the change in coordinates of the centre of mass x_T a y_T, with time, x_T(t) = ?, y_T(t) = ? respectively.

• Hint 3 – to the part A

  Assume that the balls roll down without friction. What accelerations do the balls experience? Draw a picture and mark all forces that affect the balls and find the accelerations, using the Newton’s second law. What type of motion is exhibited by single balls (in term of trajectory and velocity)?

• Solution Hint 3

  The force of gravity and the normal force of the surface act on both balls. Their resultant causes the ball of mass m₁ to accelerate in xdirection and the ball of mass m₂ to accelerate in y direction. The figure bellow shows not only the acting force of gravity F_G and normal force F_N but also the components of the force of gravity aligned with the coordinate axes (in black).

  

  According to the Newton’s second law, for the ball of mass m₁: we can derive the following equation:

  \[ \vec{F}_{G1}\,+\,\vec{F}_{N1}\,=\,m_1\vec{a}_1\tag{1}\]

  Following the picture, we rewrite the equation (1) into scalar form:

  The direction of x-axis:

  \[F_{G1}\sin{\alpha}\,=\,m_1a_1\tag{2}\]

  The direction of y-axis:

  \[F_{G1}\cos{\alpha}\,-\,F_{N1}\,=\,0\tag{3}\]

  By substitution of the force of gravity into equation (2), we obtain:

  \[m_1g\sin{\alpha}\,=\,m_1a_1\,\Rightarrow\,a_1\,=\,g\sin{\alpha}\tag{4}\]

  The ball of mass m₁ undergoes uniformly accelerated motion in the direction of the x axis and with the acceleration a₁. In the case of the second ball we proceed analogically.

  
  

  According to the Newton’s second law, for the ball of mass m₂: we can derive the following equation:

  \[ \vec{F}_{G2}\,+\,\vec{F}_{N2}\,=\,m_2\vec{a}_2\tag{5}\]

  Following the picture, we rewrite the equation (5) into scalar form:

  The direction of x-axis:

  \[F_{G2}\sin{\alpha}\,-\,F_{N2}\,=\,0\tag{6}\]

  The direction of y-axis:

  \[F_{G2}\cos{\alpha}\,=\,m_2a_2\tag{7}\]

  After substitution of the force of gravity into equation (7), we obtain:

  \[m_2g\cos{\alpha}\,=\,m_2a_2\,\Rightarrow\,a_2\,=\,g\cos{\alpha}\tag{8}\]

  The ball of mass m₂ also undergoes uniformly accelerated motion this time in the direction of the y axis and with the acceleration a₂

• Hint 4 – to the part A

  Write down expressions for x and y coordinates of the ball of mass m₁ as a function of time. Proceed the same way for the ball of mass m₂.

• Solution Hint 4

  The ball of mass m₁ undergoes uniformly accelerated motion in the direction of the x axis and with the acceleration a₁:

  \[x_1(t)\,=\,\frac{1}{2}a_1t^2\,=\,\frac{1}{2}g(\sin{\alpha})t^2\tag{9}\]

  The coordinate of the ball in the direction of y axis is identically equal to zero:

  \[y_1(t)\,=\,0\tag{10}\]

  The ball of mass m₂ undergoes uniformly accelerated motion in the direction of the y axis and with the acceleration a₂:

  \[y_2(t)\,=\,\frac{1}{2}a_2t^2\,=\,\frac{1}{2}g(\cos{\alpha})t^2\tag{11}\]

  The coordinate of the ball in the direction of x axis is identically equal to zero:

  \[x_2(t)\,=\,0\tag{12}\]

• Hint 5 – to the part A

  At this point, we already know the way of changing the coordinates of our two balls (point masses) with time. Use then the formula that determines the coordinates of the centre of mass and solve the first part of the task.

• Solution Hint 5

  The centre of mass of npoint masses are given by:

  \[\vec{r}_T\,=\,\frac{\sum_{i=1}^n{m_i\vec{r}_i}}{\sum_{i=1}^n{m_i}}\]

  where m_i are masses of individual point masses and r_i are their respective position vectors.

  Because the coordinates of the centre of mass can be calculated for each coordinate separately, after substitution from equations (9) and (10) we obtain for x-coordinate of the centre of mass of our system:

  \[x_T(t)\,=\,\frac{\sum_{i=1}^2{m_ix_i(t)}}{\sum_{i=1}^2{m_i}}\,=\,\frac{m_1x_1(t)\,+\,m_2x_2(t)} {m_1\,+\,m_2}\,=\,\frac{m_1g(\sin{\alpha})t^2}{2(m_1\,+\,m_2)}\]

  Analogous equation for y-of the centre of mass can be obtained from equations (10) and (11):

  \[y_T(t)\,=\,\frac{\sum_{i=1}^2{m_iy_i(t)}}{\sum_{i=1}^2{m_i}}\,=\,\frac{m_1y_1(t)\,+\,m_2y_2(t)} {m_1\,+\,m_2}\,=\,\frac{m_2g(\cos{\alpha})t^2}{2(m_1\,+\,m_2)}\]

  The coordinates of the centre of mass are changing with time in the chosen frame of reference as follow:

  \[x_T(t)\,=\,\frac{m_1g(\sin{\alpha})t^2}{2(m_1\,+\,m_2)}\] \[y_T(t)\,=\,\frac{m_2g(\cos{\alpha})t^2}{2(m_1\,+\,m_2)}\] \[z_T(t)\,=\,0\]

• Hint 6 – to the part B

  In the second part of the task we include the moment of inertia of balls. Firstly, go through the solution of the part A and decide how far it is replicable – where the non-zero moment of inertia is relevant for the first time?

• Solution Hint 6

  We choose a frame of reference analogously, there is only difference that the both axes are parallel with the surface of the inclined planes in distance R from it - that way we can accomplish that y-coordinate of the first ball and x-coordinate of the second ball are zero during the motion (see the figure).

  

  The non-zero moment of inertia J of balls will play its role, when we calculate an acceleration in each coordinate direction. Part of potential energy of ball E_p is converted into rotational kinetic energy E_r during the motion on inclined plane, thus less energy can be converted into translational kinetic energy of ball E_k.The ball is moving then with smaller acceleration than in the case of neglecting the moment of inertia.

• Hint 7 – to the part B

  Use the law of conservation of energy to determine an acceleration of the centre of mass of balls in each coordinate direction. (Attention, it is quite long computation, integrating pieces of knowledge from different parts of mechanics!)

• Solution 7

  We choose two situations and describe whole mechanic energy the balls have. The reference point of zero potential energy is set in depthhbelow the point of the release (see the figure above).

  For the ball of mass m₁:

  Situation 1 – moment of release:

  \[E_1\,=\,E_{p1}\tag{13}\]

  where E_p1 is potential energy of the ball of mass m₁.

  Situation 2 – the centre of mass of the ball covered a distance x:

  \[E_2\,=\,E_{r1}\,+\,E_{k1}\tag{14}\]

  where E_r1is rotational kinetic energy of the ball of mass m₁ respect to axis passing through its centre and E_k1 is translational kinetic energy.

  According the law of conservation of mechanical energy there holds:

  \[E_1\,=\,E_2\tag{15}\]

  Thus, after substitution from terms (13) and (14):

  \[E_{p1}\,=\,E_{r1}\,+\,E_{k1}\tag{16}\]

  From well-known equations for calculating the each energy we substitute in term (16):

  \[m_1gh\,=\,\frac{1}{2}m_1v^2_x\,+\,\frac{1}{2}J{\omega_1}^2\tag{17}\]

  where v_x is velocity of the centre of mass of ball, J is the moment of inertia about an axis through ball’s centre and ω₁ is its angular velocity.

  For the moment of inertia of the homogeneous ball of mass m₁ and radius R about an axis through its centre:

  \[J\,=\,\frac{2}{5}m_1R^2\tag{18}\]

  The figure shows that:

  \[\sin{\alpha}\,=\,\frac{h}{x}\,\Rightarrow\,h\,=\,x\sin{\alpha}\tag{19}\]

  The magnitude of the acceleration of the centre of mass of the first ball is denoted by a_x. By equations of uniformly accelerated linear motion we obtain:

  \[x\,=\,\frac{1}{2}a_xt^2\tag{20}\] \[v_x\,=\,a_xt\tag{21}\]

  And due to property of rotational motion with angular velocity ω₁:

  \[\omega_1\,=\,\frac{v_x}{R}\tag{22}\]

  The formulae (18) to (22) substituting into equation (17) give:

  \[m_1g{\cdot}\frac{1}{2}a_xt^2\sin{\alpha}\,=\,\frac{1}{2}m_1(a_xt)^2\,+\,\frac{1}{2}{\cdot}\frac{2}{5}m_1R^2{(\frac{a_xt}{R})^2}\tag{23}\]

  By rearranging equation (23) we obtain:

  \[a_x\,=\,\frac{5}{7}g\sin{\alpha}\tag{24}\]
  

  Analogously we can proceed also in the case of the second ball of mass m₂, only the function sine is replaced by function cosine in the calculation of the hight h.

  For the ball of mass m₂:

  The law of conservation of mechanic energy have this form:

  \[E_{p2}\,=\,E_{r2}\,+\,E_{k2}\tag{25}\]

  where E_p2 is potential energy of the ball of mass m₂, E_r2 is kinetic energy of rotational motion of the ball about an axis passing though its centre and E_k2 kinetic energy of its rectilinear motion. The potential energy is calculated in the moment of release, both kinetic energies after covering the length y(analogously as in the case of the first ball).From well-known equations for calculating the each energy we substitute in term (25):

  \[m_2gh\,=\,\frac{1}{2}m_2v^2_y\,+\,\frac{1}{2}J{\omega_2}^2\tag{26}\]

  where v_y is the velocity of the centre of mass of the ball, J is the moment of inertia about an axis through ball’s centre and a ω₂ is its angular velocity.

  The moment of inertia of the second ball we determine the same way as we did in the case of the first ball with help of formula (18) (with mass m₂), of course), the geometry of the task changes to:

  \[\cos{\alpha}\,=\,\frac{h}{y}\,\Rightarrow\,h\,=\,y\cos{\alpha}\tag{27}\]

  A magnitude of acceleration of the centre of mass of the second ball is denotes by a_y. By equations of uniformly accelerated linear motion we obtain:

  \[y\,=\,\frac{1}{2}a_yt^2\tag{28}\] \[v_y\,=\,a_yt\tag{29}\]

  And due to property of rotational motion with angular velocity ω₂:

  \[\omega_2\,=\,\frac{v_y}{R}\tag{30}\]

  formulae (27) to (30) substituting into equation (26) give:

  \[m_2g{\cdot}\frac{1}{2}a_yt^2\cos{\alpha}\,=\,\frac{1}{2}m_2(a_yt)^2\,+\,\frac{1}{2}{\cdot}\frac{2}{5}m_2R^2{(\frac{a_yt}{R})^2}\tag{31}\]

  By rearranging equation (31) we obtain:

  \[a_y\,=\,\frac{5}{7}g\cos{\alpha}\tag{32}\]

• Commentary 1

  A magnitude of acceleration of the centre of mass of both balls can be determined also with different technique, and it is with the help of the laws of conservation of momentum and angular momentum.

• Hint 8 – to the part B

  In previous step we obtained the magnitude of acceleration of the centre of the first ball a_x and the second ball a_y. Use them as components a₁ and a₂ in the part A and determine the change of coordinates x_T´(t), y_T´(t) of the centre of mass of the system with time using the same computation method.

• Solution 8

  In the next section we repeat the same technique as described in hints (5) and (6), only with different magnitudes of accelerations:

  The centre of mass of the ball of mass m₁ undergoes uniformly accelerated motion in the direction of the x axis and with the acceleration a_x:

  \[x_1´(t)\,=\,\frac{1}{2}a_xt^2\,=\,\frac{5}{14}g(\sin{\alpha})t^2\]

  The coordinate of the ball in the direction of y axis is identically equal to zero:

  \[y_1´(t)\,=\,0\]

  The centre of mass of the ball of mass m₂ uniformly accelerated motion in the direction of the y axis and with the acceleration a_y:

  \[y_2´(t)\,=\,\frac{1}{2}a_yt^2\,=\,\frac{5}{14}g(\cos{\alpha})t^2\]

  The coordinate of the ball in the direction of x axis is identically equal to zero:

  \[x_2´(t)\,=\,0\]

  The coordinates of the centre of mass of the system can be calculated for each coordinate separately:

  \[x_T´(t)\,=\,\frac{\sum_{i=1}^2{m_ix_i´(t)}}{\sum_{i=1}^2{m_i}}\,=\,\frac{m_1x_1´(t)\,+\,m_2x_2´(t)} {m_1\,+\,m_2}\,=\,\frac{5m_1g(\sin{\alpha})t^2}{14(m_1\,+\,m_2)}\] \[y_T´(t)\,=\,\frac{\sum_{i=1}^2{m_iy_i´(t)}}{\sum_{i=1}^2{m_i}}\,=\,\frac{m_1y_1´(t)\,+\,m_2y_2´(t)} {m_1\,+\,m_2}\,=\,\frac{5m_2g(\cos{\alpha})t^2}{14(m_1\,+\,m_2)}\]

  The coordinates of the centre of mass are changing with time in the chosen frame of reference as follow:

  \[x_T´(t)\,=\,\frac{5m_1g(\sin{\alpha})t^2}{14(m_1\,+\,m_2)}\] \[y_T´(t)\,=\,\frac{5m_2g(\cos{\alpha})t^2}{14(m_1\,+\,m_2)}\] \[z_T´(t)\,=\,0\]

• Solution - the part A

  If we neglect the moment of inertia of balls, it is possible to replace balls with point masses. The figure bellow shows the appropriate reference frame used in this problem.

  

  We consider that the motion is situated in the plane xy, thus we can omit the change of z-coordinate. The output of the task should be a formula which describes the change in coordinates of the centre of mass x_T a y_T, with time, x_T(t) = ?, y_T(t) = ? respectively.

  The force of gravity and the normal force of the surface act on both balls. Their resultant causes the ball of mass m₁ to accelerate in xdirection and the ball of mass m₂ to accelerate in y direction. The figure bellow shows not only the acting force of gravity F_G and normal force F_N but also the components of the force of gravity aligned with the coordinate axes (in black).

  

  According to the Newton’s second law, for the ball of mass m₁: we can derive the following equation:

  \[ \vec{F}_{G1}\,+\,\vec{F}_{N1}\,=\,m_1\vec{a}_1\tag{1}\]

  Following the picture, we rewrite the equation (1) into scalar form:

  The direction of x-axis:

  \[F_{G1}\sin{\alpha}\,=\,m_1a_1\tag{2}\]

  The direction of y-axis:

  \[F_{G1}\cos{\alpha}\,-\,F_{N1}\,=\,0\tag{3}\]

  By substitution of the force of gravity into equation (2), we obtain:

  \[m_1g\sin{\alpha}\,=\,m_1a_1\,\Rightarrow\,a_1\,=\,g\sin{\alpha}\tag{4}\]

  The ball of mass m₁ undergoes uniformly accelerated motion in the direction of the x axis and with the acceleration a₁. In the case of the second ball we proceed analogically.

  
  

  According to the Newton’s second law, for the ball of mass m₂: we can derive the following equation:

  \[ \vec{F}_{G2}\,+\,\vec{F}_{N2}\,=\,m_2\vec{a}_2\tag{5}\]

  Following the picture, we rewrite the equation (5) into scalar form:

  The direction of x-axis:

  \[F_{G2}\sin{\alpha}\,-\,F_{N2}\,=\,0\tag{6}\]

  The direction of y-axis:

  \[F_{G2}\cos{\alpha}\,=\,m_2a_2\tag{7}\]

  After substitution of the force of gravity into equation (7), we obtain:

  \[m_2g\cos{\alpha}\,=\,m_2a_2\,\Rightarrow\,a_2\,=\,g\cos{\alpha}\tag{8}\]

  The ball of mass m₂ also undergoes uniformly accelerated motion this time in the direction of the y axis and with the acceleration a₂

  
  

  The ball of mass m₁ undergoes uniformly accelerated motion in the direction of the x axis and with the acceleration a₁:

  \[x_1(t)\,=\,\frac{1}{2}a_1t^2\,=\,\frac{1}{2}g(\sin{\alpha})t^2\tag{9}\]

  The coordinate of the ball in the direction of y axis is identically equal to zero:

  \[y_1(t)\,=\,0\tag{10}\]

  The ball of mass m₂ undergoes uniformly accelerated motion in the direction of the y axis and with the acceleration a₂:

  \[y_2(t)\,=\,\frac{1}{2}a_2t^2\,=\,\frac{1}{2}g(\cos{\alpha})t^2\tag{11}\]

  The coordinate of the ball in the direction of x axis is identically equal to zero:

  \[x_2(t)\,=\,0\tag{12}\]

  The centre of mass of npoint masses are given by:

  \[\vec{r}_T\,=\,\frac{\sum_{i=1}^n{m_i\vec{r}_i}}{\sum_{i=1}^n{m_i}}\]

  where m_i are masses of individual point masses and r_i are their respective position vectors.

  Because the coordinates of the centre of mass can be calculated for each coordinate separately, after substitution from equations (9) and (10) we obtain for x-coordinate of the centre of mass of our system:

  \[x_T(t)\,=\,\frac{\sum_{i=1}^2{m_ix_i(t)}}{\sum_{i=1}^2{m_i}}\,=\,\frac{m_1x_1(t)\,+\,m_2x_2(t)} {m_1\,+\,m_2}\,=\,\frac{m_1g(\sin{\alpha})t^2}{2(m_1\,+\,m_2)}\]

  Analogous equation for y-of the centre of mass can be obtained from equations (10) and (11):

  \[y_T(t)\,=\,\frac{\sum_{i=1}^2{m_iy_i(t)}}{\sum_{i=1}^2{m_i}}\,=\,\frac{m_1y_1(t)\,+\,m_2y_2(t)} {m_1\,+\,m_2}\,=\,\frac{m_2g(\cos{\alpha})t^2}{2(m_1\,+\,m_2)}\]

  The coordinates of the centre of mass are changing with time in the chosen frame of reference as follow:

  \[x_T(t)\,=\,\frac{m_1g(\sin{\alpha})t^2}{2(m_1\,+\,m_2)}\] \[y_T(t)\,=\,\frac{m_2g(\cos{\alpha})t^2}{2(m_1\,+\,m_2)}\] \[z_T(t)\,=\,0\]

• Solution - the part B

  We choose a frame of reference analogously, there is only difference that the both axes are parallel with the surface of the inclined planes in distance R from it - that way we can accomplish that y-coordinate of the first ball and x-coordinate of the second ball are zero during the motion (see the figure).

  

  The non-zero moment of inertia J of balls will play its role, when we calculate an acceleration in each coordinate direction. Part of potential energy of ball E_p is converted into rotational kinetic energy E_r during the motion on inclined plane, thus less energy can be converted into translational kinetic energy of ball E_k.The ball is moving then with smaller acceleration than in the case of neglecting the moment of inertia.

  Use the law of conservation of energy to determine an acceleration of the centre of mass of balls in each coordinate direction.

  We choose two situations and describe whole mechanic energy the balls have. The reference point of zero potential energy is set in depthhbelow the point of the release (see the figure above).

  For the ball of mass m₁:

  Situation 1 – moment of release:

  \[E_1\,=\,E_{p1}\tag{13}\]

  where E_p1 is potential energy of the ball of mass m₁.

  Situation 2 – the centre of mass of the ball covered a distance x:

  \[E_2\,=\,E_{r1}\,+\,E_{k1}\tag{14}\]

  where E_r1is rotational kinetic energy of the ball of mass m₁ respect to axis passing through its centre and E_k1 is translational kinetic energy.

  According the law of conservation of mechanical energy there holds:

  \[E_1\,=\,E_2\tag{15}\]

  Thus, after substitution from terms (13) and (14):

  \[E_{p1}\,=\,E_{r1}\,+\,E_{k1}\tag{16}\]

  From well-known equations for calculating the each energy we substitute in term (16):

  \[m_1gh\,=\,\frac{1}{2}m_1v^2_x\,+\,\frac{1}{2}J{\omega_1}^2\tag{17}\]

  where v_x is velocity of the centre of mass of ball, J is the moment of inertia about an axis through ball’s centre and ω₁ is its angular velocity.

  For the moment of inertia of the homogeneous ball of mass m₁ and radius R about an axis through its centre:

  \[J\,=\,\frac{2}{5}m_1R^2\tag{18}\]

  The figure shows that:

  \[\sin{\alpha}\,=\,\frac{h}{x}\,\Rightarrow\,h\,=\,x\sin{\alpha}\tag{19}\]

  The magnitude of the acceleration of the centre of mass of the first ball is denoted by a_x. By equations of uniformly accelerated linear motion we obtain:

  \[x\,=\,\frac{1}{2}a_xt^2\tag{20}\] \[v_x\,=\,a_xt\tag{21}\]

  And due to property of rotational motion with angular velocity ω₁:

  \[\omega_1\,=\,\frac{v_x}{R}\tag{22}\]

  The formulae (18) to (22) substituting into equation (17) give:

  \[m_1g{\cdot}\frac{1}{2}a_xt^2\sin{\alpha}\,=\,\frac{1}{2}m_1(a_xt)^2\,+\,\frac{1}{2}{\cdot}\frac{2}{5}m_1R^2{(\frac{a_xt}{R})^2}\tag{23}\]

  By rearranging equation (23) we obtain:

  \[a_x\,=\,\frac{5}{7}g\sin{\alpha}\tag{24}\]
  

  Analogously we can proceed also in the case of the second ball of mass m₂, only the function sine is replaced by function cosine in the calculation of the hight h.

  For the ball of mass m₂:

  The law of conservation of mechanic energy have this form:

  \[E_{p2}\,=\,E_{r2}\,+\,E_{k2}\tag{25}\]

  where E_p2 is potential energy of the ball of mass m₂, E_r2 is kinetic energy of rotational motion of the ball about an axis passing though its centre and E_k2 kinetic energy of its rectilinear motion. The potential energy is calculated in the moment of release, both kinetic energies after covering the length y(analogously as in the case of the first ball).From well-known equations for calculating the each energy we substitute in term (25):

  \[m_2gh\,=\,\frac{1}{2}m_2v^2_y\,+\,\frac{1}{2}J{\omega_2}^2\tag{26}\]

  where v_y is the velocity of the centre of mass of the ball, J is the moment of inertia about an axis through ball’s centre and a ω₂ is its angular velocity.

  The moment of inertia of the second ball we determine the same way as we did in the case of the first ball with help of formula (18) (with mass m₂), of course), the geometry of the task changes to:

  \[\cos{\alpha}\,=\,\frac{h}{y}\,\Rightarrow\,h\,=\,y\cos{\alpha}\tag{27}\]

  A magnitude of acceleration of the centre of mass of the second ball is denotes by a_y. By equations of uniformly accelerated linear motion we obtain:

  \[y\,=\,\frac{1}{2}a_yt^2\tag{28}\] \[v_y\,=\,a_yt\tag{29}\]

  And due to property of rotational motion with angular velocity ω₂:

  \[\omega_2\,=\,\frac{v_y}{R}\tag{30}\]

  formulae (27) to (30) substituting into equation (26) give:

  \[m_2g{\cdot}\frac{1}{2}a_yt^2\cos{\alpha}\,=\,\frac{1}{2}m_2(a_yt)^2\,+\,\frac{1}{2}{\cdot}\frac{2}{5}m_2R^2{(\frac{a_yt}{R})^2}\tag{31}\]

  By rearranging equation (31) we obtain:

  \[a_y\,=\,\frac{5}{7}g\cos{\alpha}\tag{32}\]
  

  In the next section we repeat the same technique as described in hints (5) and (6), only with different magnitudes of accelerations:

  The centre of mass of the ball of mass m₁ undergoes uniformly accelerated motion in the direction of the x axis and with the acceleration a_x:

  \[x_1´(t)\,=\,\frac{1}{2}a_xt^2\,=\,\frac{5}{14}g(\sin{\alpha})t^2\]

  The coordinate of the ball in the direction of y axis is identically equal to zero:

  \[y_1´(t)\,=\,0\]

  The centre of mass of the ball of mass m₂ uniformly accelerated motion in the direction of the y axis and with the acceleration a_y:

  \[y_2´(t)\,=\,\frac{1}{2}a_yt^2\,=\,\frac{5}{14}g(\cos{\alpha})t^2\]

  The coordinate of the ball in the direction of x axis is identically equal to zero:

  \[x_2´(t)\,=\,0\]

  The coordinates of the centre of mass of the system can be calculated for each coordinate separately:

  \[x_T´(t)\,=\,\frac{\sum_{i=1}^2{m_ix_i´(t)}}{\sum_{i=1}^2{m_i}}\,=\,\frac{m_1x_1´(t)\,+\,m_2x_2´(t)} {m_1\,+\,m_2}\,=\,\frac{5m_1g(\sin{\alpha})t^2}{14(m_1\,+\,m_2)}\] \[y_T´(t)\,=\,\frac{\sum_{i=1}^2{m_iy_i´(t)}}{\sum_{i=1}^2{m_i}}\,=\,\frac{m_1y_1´(t)\,+\,m_2y_2´(t)} {m_1\,+\,m_2}\,=\,\frac{5m_2g(\cos{\alpha})t^2}{14(m_1\,+\,m_2)}\]

  The coordinates of the centre of mass are changing with time in the chosen frame of reference as follow:

  \[x_T´(t)\,=\,\frac{5m_1g(\sin{\alpha})t^2}{14(m_1\,+\,m_2)}\] \[y_T´(t)\,=\,\frac{5m_2g(\cos{\alpha})t^2}{14(m_1\,+\,m_2)}\] \[z_T´(t)\,=\,0\]

• Answer

  Part A: (We neglect the moment of inertia of balls.)

  The coordinates of the centre of mass are changing with time in the chosen frame of reference as follow:

  \[x_T(t)\,=\,\frac{m_1g(\sin{\alpha})t^2}{2(m_1\,+\,m_2)}\] \[y_T(t)\,=\,\frac{m_2g(\cos{\alpha})t^2}{2(m_1\,+\,m_2)}\] \[z_T(t)\,=\,konst.\]
  

  Part B: (We include the moment of inertia of balls.)

  The coordinates of the centre of mass are changing with time in the chosen frame of reference as follow:

  \[x_T´(t)\,=\,\frac{5m_1g(\sin{\alpha})t^2}{14(m_1\,+\,m_2)}\] \[y_T´(t)\,=\,\frac{5m_2g(\cos{\alpha})t^2}{14(m_1\,+\,m_2)}\] \[z_T´(t)\,=\,konst.\]

• Commentary 2

  The solution makes sense only for such t until the one of the balls touches the horizontal surface.