Question

A ball of mass $20g$ hits a smooth wall at an angle of ${45^ \circ }$ with a velocity of $15m/s$. If the ball rebounds at ${90^ \circ }$ to the direction of incidence, calculate the impulse received by the ball.

Answer 1

This question can be solved by using concepts of Newton’s second law of motion. Impulse is the change in momentum and can be found by assigning the values to the equation.
Formula Used: The formulae used in the solution are given here.
Impulse $I = m\Delta v$ where mass of the ball is $m$ and change in velocity is $\Delta v$.

Complete Step by Step Solution: Impulse is the integral of a force, over the time interval, for which it acts.
Impulse is a term that quantifies the overall effect of a force acting over time. It is conventionally given the symbol $J$ and expressed in Newton-seconds.
For a constant force, $J = F \cdot \Delta t$.
As we saw earlier, this is exactly equivalent to a change in momentum $\Delta p$. This equivalence is known as the impulse-momentum theorem. Because of the impulse-momentum theorem, we can make a direct connection between how a force acts on an object over time and the motion of the object.
It has been given that a ball of mass $20g$ hits a smooth wall at an angle of ${45^ \circ }$ with a velocity of $15m/s$. The ball rebounds at ${90^ \circ }$ to the direction of incidence.
The mass of the ball is $m = 20g = 0.02kg$ and velocity $v = 15m/s$.
Since the change in momentum is impulse, we have impulse = $m\Delta v = m\left( {{v_f} - {v_i}} \right)\sin \theta$.
Substituting the values in the equation, we have, $m\left( {{v_f} - {v_i}} \right)\sin \theta = 0.02\left( { - 15 - 15} \right)\sin {45^ \circ }$.
Simplifying the equation, we have, $0.02 \times 30 \times \dfrac{1}{{\sqrt 2 }} = 0.424Ns$.

Note: Since force is a vector quantity, impulse is also a vector quantity. Impulse applied to an object produces an equivalent vector change in its linear momentum, also in the same direction.
For objects and systems with constant mass, Newton’s second law of motion can be stated in terms of an object's acceleration.
$F = m\dfrac{{dv}}{{dt}}$
Rearranging the equation, we can write, $Fdt = mdv$.
Integrating the equation on both the side,
$F\int {dt} = m\int {dv}$
$\Rightarrow F \cdot t = mv$
This product of the force and time is called impulse.