Question

A ball of mass 'm' moves normal to a wall with a velocity 'u' and rebounds with a velocity 'v'. The change in momentum of the ball is (surface of the wall is smooth):

• A. $m(u+v)$ towards the wall
• B. $m(u+v)$ away from the wall
• C. $m(u-v)$ towards the wall
• D. $m(u-v)$ away from the wall

Answer 1

Answer: (B)

A ball of mass, m, moving with initial velocity, u to the right towards a wall.
It will have momentum $\overrightarrow{p_i}=m\overrightarrow{u}$ towards the right.

The ball bounces off the wall. It will now be moving to the left, with the same mass, but a different velocity, v and therefore, a different momentum,$\overrightarrow{p_f}=m\overrightarrow{v}$ towards the left.

The final momentum vector must be the sum of the initial momentum vector and the change in momentum vector, $△\overrightarrow{p}=m\overrightarrow{△v}$ .

Using tail to head vector addition, $△\overrightarrow{p}$, must be the vector that starts at the head of $\overrightarrow{p_i}$ and ends on the head of $\overrightarrow{p_f}$, hence the resultant change in momentum vector will point towards the left, that is, away from the wall.

We also know from algebraic addition of vectors that:
$\overrightarrow{p_f}=\overrightarrow{p_i}+△\overrightarrow{p}$

$\overrightarrow{p_f}-\overrightarrow{p_i}=△\overrightarrow{p}$

Magnitude of the change in momentum = $mv-(-u)=m(v+u)$
Direction of change in momentum is towards the left, away from the wall.