Question

A ball of mass m moving with a speed of v towards a wall gets rebound with the same speed backward.what is the impulse imparted on the ball?

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

The ball is thrown at a velocity "v".

$\huge{\bold{\underline{Explanation:-}}}$

After the ball rebounds the velocity is same but in opposite direction.

So, the Speed at which the ball is thrown "v" and it returns with a speed "- v"

(Here negative sign indicates the direction is opposite to the Velocity in the first case.)

Now, Impulse

$\large{\bold{I = m(v - u)}}$

As Impulse is equal to change in momentum,

$\large{I = m( - v - v)}$

$\large{I = m \times -2v}$

$\huge{\boxed{\boxed{I = - 2mv}}}$

So, the Impulse imparted on the ball is - 2mv.

Additional formula :-

• $\large{\bold{I = Force \times Displacement}}$

• Area under Force displacement graph gives Impulse and change in momentum.

Here "I" denotes Impulse acting on the body.

Answer 2

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

From the Question,

• Mass of the Ball = m

ATQ,

The Initial Velocity of the ball is equal to the velocity of the object after rebound

Imaging a Cartesian Plane,the Initial Velocity would be "v" and the final velocity would be "- v",because after rebound the ball's motion is towards that of negative x - axis

We Know that,

$\large{ \sf{lmpulse = change \: in \: momentum}}$

Now,

$\sf{l = mv \: - \: mu} \\ \\ \huge{ \rightarrow \: \underline{ \boxed{ \sf{l = m(v - u)}}}}$

Putting the values,we get:

$\sf{l = m[ - v \: + \: ( - v)] } \\ \\ \implies \: \huge{\boxed{ \boxed{\sf{l = - 2mv}}}}$

Therefore,the impulse imparted on the ball is -2mv N.s