Question

A ball of mass m moving with velocity v_i_vec strikes a vertical wall. The angle between the ball's initial velocity vector and the wall is theta_i as shown on the diagram, which depicts the situation as seen from above. The duration of the collision between the ball and the wall is Deltat, and this collision is completely elastic. Friction is negligible, so the ball does not start spinning. In this idealized collision, the force exerted on the ball by the wall is parallel to the x axis.

A ball of mass m moving with velocity v_i_vec strikes a vertical wall. The angle between the ball's initial velocity vector and the wall is theta_i as shown on the diagram, which depicts the situation as seen from above. The duration of the collision between the ball and the wall is Deltat, and this collision is completely elastic. Friction is negligible, so the ball does not start spinning. In this idealized collision, the force exerted on the ball by the wall is parallel to the x axis. Part A:What is the final angle theta f that the ball's velocity vector makes with the negative y axis?Part B: What is the magnitude F of the average force exerted on the ball by the wall?

Part A:What is the final angle $theta_f$ that the ball's velocity vector makes with the negative y axis?

Part B: What is the magnitude of the average force exerted on the ball by the wall?

Answer 1

Concepts and reason

This question is based on the concepts of force, momentum and elastic collision.

First resolve the initial velocity of ball in the x and y components. Then, calculate the final velocity of the ball and its angle.

Fundamentals

Force-$F$ on an object is defined as the rate of change of its momentum. This can be expressed as follows:

$F = \frac{{\Delta p}}{{\Delta t}}$

Here, $\Delta p$ is the change in momentum of the object in time$\Delta t$.

Momentum-$p$ of an object is the product of its mass $m$ and velocity $v$. This is expressed as follows:

$p = mv$

Magnitude of a vector $\vec v$ with component ${v_x}$ and ${v_y}$ is given by relation;

$\left| {\vec v} \right| = \sqrt {{v_x}^2 + {v_y}^2}$

When two objects collide, transfer of energy and momentum takes place. The wall during collision imparts force on the wall and changes its momentum. The energy of the system (wall and ball) remains conserved during the collision because the collision is elastic.

In the perfectly elastic collision, if two objects of masses ${m_1}$ and ${m_2}$with initial velocities ${v_1}$ and ${v_2}$ collide, their final velocities ${v_1}^\prime$ and ${v_2}^\prime$ are expressed as follows:

${v_1}^\prime = \frac{{{m_1} - {m_2}}}{{{m_1} + {m_2}}}{v_1} + \frac{{2{m_2}}}{{{m_1} + {m_2}}}{v_2}$ ……(1)

${v_2}^\prime = \frac{{2{m_1}}}{{{m_1} + {m_2}}}{v_1} - \frac{{{m_1} - {m_2}}}{{{m_1} + {m_2}}}{v_2}$ ……(2)

Calculate x-component of initial velocity ${v_i}$ using trigonometric relations as follows:

${v_x} = {v_i}\sin {\theta _i}$

Similarly, calculate y-component of initial velocity as follows:

${v_y} = {v_i}\cos {\theta _i}$

Calculate final velocities along x-axis and y-axis by using the relationship (1).

Calculate final velocities along x-axis as follows:

Velocity of object 1 after collision is given by (1).

${v_1}^\prime = \frac{{{m_1} - {m_2}}}{{{m_1} + {m_2}}}{v_1} + \frac{{2{m_2}}}{{{m_1} + {m_2}}}{v_2}$

Substitute $m$ for ${m_1}$, $M$ for ${m_2}$, $v\sin {\theta _i}$ for ${v_1}$ and $0$ for ${v_2}$ in the equation (1) to calculate the horizontal component of final velocity.

$\begin{array}{c}\\{v_x}^\prime = \frac{{m - M}}{{m + M}}v\sin {\theta _i} + \frac{{2M}}{{m + M}}\left( 0 \right)\\\\ = \frac{{m - M}}{{m + M}}v\sin {\theta _i}\\\end{array}$

Since mass of wall is much greater than the mass of the ball.

Substitute $0$ for $m$ in the above equation of ${v_x}^\prime$.

$\begin{array}{c}\\{v_x}^\prime = \frac{{0 - M}}{{0 + M}}v\sin {\theta _i}\\\\ = \frac{{ - M}}{M}v\sin {\theta _i}\\\\ = - v\sin {\theta _i}\\\end{array}$

Only x-component of velocity is involved. Therefore y component of velocity that is ${v_y}$ remains unchanged. This can be expressed as follows:

$\begin{array}{c}\\{v_y}^\prime = {v_y}\\\\ = v\cos {\theta _i}\\\end{array}$

Part A

Calculate the angle of the final velocity as follows:

$\tan {\theta _f} = \frac{{{{v'}_x}}}{{{{v'}_y}}}$

Substitute $- v\sin {\theta _i}$ for ${v'_x}$ and $v\cos \theta$ for ${v'_y}$ in the equation ${\theta _f} = {\tan ^{ - 1}}\left( {\frac{{{{v'}_x}}}{{{{v'}_y}}}} \right)$ to calculate angle of the final velocity.

$\begin{array}{c}\\{\theta _f} = {\tan ^{ - 1}}\left( {\frac{{ - v\sin {\theta _i}}}{{v\cos {\theta _i}}}} \right)\\\\ = {\tan ^{ - 1}}\left( { - \tan {\theta _i}} \right)\\\\ = {\tan ^{ - 1}}\left( {\tan \left( { - {\theta _i}} \right)} \right)\\\\ = - {\theta _i}\\\end{array}$

Here negative sign shows that angle measured is anticlockwise.

Hence, the angle of final velocity is ${\theta _i}$ from negative $y$ axis.

Calculate change in momentum along x-axis as follows:

Substitute $- v\sin {\theta _i}$ for ${v_x}^\prime$ and $v\sin {\theta _i}$ for${v_x}$ in the equation $\Delta {p_x} = m{v_x}^\prime - m{v_x}$to calculate change in momentum along x-axis.

$\begin{array}{c}\\\Delta {p_x} = m\left( { - v\sin {\theta _i}} \right) - m\left( {v\sin {\theta _i}} \right)\\\\ = - mv\sin {\theta _i} - mv\sin \theta \\\\ = - 2mv\sin {\theta _i}\\\end{array}$

No collision takes place in y-direction. y-component of momentum doesn’t change. This can be expressed as follows:

$\Delta {p_y} = 0$

Calculate force ${F_x}$ along x-axis by using definition of force as follows:

Substitute $- 2mv\sin {\theta _i}$ for $\Delta {p_x}$ in the equation ${F_x} = \frac{{\Delta {p_x}}}{{\Delta t}}$.

${F_x} = \frac{{ - 2mv\sin {\theta _i}}}{{\Delta t}}$

Calculate force ${F_y}$ along y-axis by using definition of force as follows:

Substitute $0$ for $\Delta {p_y}$ in the equation ${F_y} = \frac{{\Delta {p_y}}}{{\Delta t}}$.

$\begin{array}{c}\\{F_x} = \frac{0}{{\Delta t}}\\\\ = 0\\\end{array}$

Part B

Calculate magnitude of the force by using relation of magnitude of a vector as follows:

Substitute $\frac{{ - 2mv\sin {\theta _i}}}{{\Delta t}}$ for ${F_x}$ and $0$ for ${F_y}$ in the equation $\left| {\vec F} \right| = \sqrt {{F_x}^2 + {F_y}^2}$ to calculate the magnitude of force.

$\begin{array}{c}\\\left| {\vec F} \right| = \sqrt {{{\left( {\frac{{ - 2mv\sin {\theta _i}}}{{\Delta t}}} \right)}^2} + {{\left( 0 \right)}^2}} \\\\ = \sqrt {{{\left( {\frac{{ - 2mv\sin {\theta _i}}}{{\Delta t}}} \right)}^2}} \\\\ = \frac{{2mv\sin {\theta _i}}}{{\Delta t}}\\\end{array}$

Hence, the magnitude of force is $\frac{{2mv\sin {\theta _i}}}{{\Delta t}}$.

Ans: Part A

The angle of final velocity vector is ${\theta _i}$.