Question

A ball of mass \(0.5\ \text{kg}\) moving horizontally at \(10\ \text{m s}^{-1}\) strikes a vertical wall and rebounds with speed \(V\). The magnitude of the change in linear momentum is found to be \(8.0\ \text{kg m s}^{-1}\). The magnitude of \(V\) is

• A. \(6.0\ \text{m s}^{-1}\)
• B. \(9.0\ \text{m s}^{-1}\)
• C. \(26.0\ \text{m s}^{-1}\)
• D. \(13.0\ \text{m s}^{-1}\)

Hint:

When an object rebounds, its velocity changes direction. So take the final velocity with the opposite sign while calculating change in momentum.

Answer 1

The Correct Option is A

Step 1: Write the given values.
Mass of the ball is
\[ m=0.5\ \text{kg} \]
Initial velocity is
\[ u=10\ \text{m s}^{-1} \]
After striking the wall, the ball rebounds in the opposite direction with speed \(V\).
So, final velocity is
\[ v=-V \]

Step 2: Use change in momentum.
Change in momentum is
\[ \Delta p=m(v-u) \]
Substituting values,
\[ \Delta p=0.5(-V-10) \]
The magnitude is given as \(8.0\ \text{kg m s}^{-1}\), so
\[ |\Delta p|=8 \]
\[ 0.5(V+10)=8 \]

Step 3: Solve for \(V\).
\[ V+10=16 \]
\[ V=6 \]

Step 4: Final conclusion.
Hence, the rebound speed is
\[ \boxed{6.0\ \text{m s}^{-1}} \]