A ping - pong ball strikes a wall with a velocity of $10m{s^{ - 1}}$. If the collision is perfectly elastic. Find the velocity of the ball after impact?
A) $-20m{s^{ - 1}}$
B) $-5m{s^{ - 1}}$
C) $1m{s^{ - 1}}$
D) $-10m{s^{ - 1}}$
Answer
249.9k+ views
Hint: To solve this question we have to use the formulae of conservation of linear momentum. Here perfectly elastic collision is taking place. Hence the total momentum before and after collision will be equal. The wall will have no velocity as it will be stationary.
Formulae used:
$e = \dfrac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}$
Here ${v_2}$ is the final velocity of the second body, ${v_1}$ is the final velocity of the first body , $e$ is the coefficient of elasticity, ${u_2}$ is the initial velocity of the second body, ${u_2}$ is the initial velocity of the first body.
Complete step by step answer:

In the question it's said that a ball moving with velocity $10m{s^{ - 1}}$ collides with a wall. We can say that,
${u_b} = 10m{s^{ - 1}}$ and ${u_w} = 0$
Here ${u_b}$ and ${u_w}$ are the initial velocities of ball and wall respectively.
Let the velocities after collision be ${v_b}$ and ${v_w}$ of the ball and wall respectively.
It is said that the collision is perfectly elastic. So the coefficient of elasticity will be
$ \Rightarrow e = 1$
Applying the conservation of momentum for a perfectly elastic, we get
$e = \dfrac{{{v_w} - {v_b}}}{{{u_b} - {u_w}}}$
Here ${v_w}$ is the final velocity of the wall, ${v_b}$ is the final velocity of the ball, $e$ is the coefficient of elasticity, ${u_w}$ is the initial velocity of the wall body, ${u_b}$ is the initial velocity of the ball.
Substituting the value of
${u_b} = 10m{s^{ - 1}}$ ,${u_w} = 0$, ${v_w} = 0$ and $e = 1$
We get,
$ \Rightarrow 1 = \dfrac{{0 - {v_b}}}{{10 - 0}}$
$ \Rightarrow {v_b} = - 10$
Hence the final velocity of the ball will be $ - 10m{s^{ - 1}}$.
This means the ball will move in the opposition direction of its initial direction.
Hence option (D) is the correct answer.
Note: Collision are of three types:
1. Perfectly elastic collision- The momentum and energy is conserved before and after collision.
2. Inelastic collision- The momentum and energy is conserved before and after collision but the total kinetic energy is not conserved after collision.
Formulae used:
$e = \dfrac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}$
Here ${v_2}$ is the final velocity of the second body, ${v_1}$ is the final velocity of the first body , $e$ is the coefficient of elasticity, ${u_2}$ is the initial velocity of the second body, ${u_2}$ is the initial velocity of the first body.
Complete step by step answer:

In the question it's said that a ball moving with velocity $10m{s^{ - 1}}$ collides with a wall. We can say that,
${u_b} = 10m{s^{ - 1}}$ and ${u_w} = 0$
Here ${u_b}$ and ${u_w}$ are the initial velocities of ball and wall respectively.
Let the velocities after collision be ${v_b}$ and ${v_w}$ of the ball and wall respectively.
It is said that the collision is perfectly elastic. So the coefficient of elasticity will be
$ \Rightarrow e = 1$
Applying the conservation of momentum for a perfectly elastic, we get
$e = \dfrac{{{v_w} - {v_b}}}{{{u_b} - {u_w}}}$
Here ${v_w}$ is the final velocity of the wall, ${v_b}$ is the final velocity of the ball, $e$ is the coefficient of elasticity, ${u_w}$ is the initial velocity of the wall body, ${u_b}$ is the initial velocity of the ball.
Substituting the value of
${u_b} = 10m{s^{ - 1}}$ ,${u_w} = 0$, ${v_w} = 0$ and $e = 1$
We get,
$ \Rightarrow 1 = \dfrac{{0 - {v_b}}}{{10 - 0}}$
$ \Rightarrow {v_b} = - 10$
Hence the final velocity of the ball will be $ - 10m{s^{ - 1}}$.
This means the ball will move in the opposition direction of its initial direction.
Hence option (D) is the correct answer.
Note: Collision are of three types:
1. Perfectly elastic collision- The momentum and energy is conserved before and after collision.
2. Inelastic collision- The momentum and energy is conserved before and after collision but the total kinetic energy is not conserved after collision.
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