Question

A ball of mass 0.5 Kg moving with a velocity of \[2m{s^{ - 1}}\]strikes a wall normally and bounces back with the same speed. If the time of contact between the ball and the wall is one millisecond, the average force exerted by the wall on the ball is:
A. 2000 N
B. 1000 N
C. 5000 N
D. 125 N

Answer 1

Hint: As the ball changes its direction after striking the wall. So, we need to show the change in direction by a negative sign. Also, the velocity of the ball changes due to this, there must be a change in momentum which is equal to the force exerted by the wall.



Formula used:
The force is defined as the rate of change of linear momentum and written as:
\[F = \dfrac{{mv - mu}}{t}\]
Where m is the mass of an object
v is final velocity
u is initial velocity
t is time taken




Complete answer:
 Given mass of a ball, m = 0.5 Kg
Final velocity of a ball, v = \[2m{s^{ - 1}}\]
Initial velocity of a ball, v = \[ - 2m{s^{ - 1}}\]( negative sign shows the direction of the velocity changes after striking the wall)
Time, t = 1 ms = \[{10^{ - 3}}\] sec
As we know that,
\[F = \dfrac{{mv - mu}}{t}\]
By using the given values, we get
\[F = \dfrac{{0.5 \times 2 - 0.5 \times ( - 2)}}{{{{10}^{ - 3}}}}\]
\[F = \dfrac{{1 + 1}}{{{{10}^{ - 3}}}}\]
\[F = 2 \times {10^3}\]
\[F = 2000{\rm{ N}}\]
Therefore, the average force exerted by the wall on the ball is 2000 N.
Hence option A is the correct answer




Note: From Newton’s second law of motion, force is equal to the time rate of change of momentum. By differentiating the momentum with respect to the time we can get the value for the force. The term momentum is defined as the product of mass and the velocity of an object.