Question

A ball of mass 1 kg moving with a velocity of 8 metre per second rebounds after striking a wall calculate the change in momentum if the ball rebounds with the same speed also calculate the force if time of contact with wall is 0.1 second​

Answer 1

GIVEN :-

• Mass of ball ( m ) = 1 kg.
• Final velocity ( v ) = - 8 m/s [ Rebounds after striking the wall ]
• Initial velocity ( u ) = 8 m/s.
• Time ( T ) = 0.1 sec.

TO FIND :-

• The change in momentum.
• The Force applied by the wall.

SOLUTION :-

As we know that change in momentum is given by,

$\implies \displaystyle \sf \: \Delta p = mv - mu$

$\implies \displaystyle \sf \: \Delta p =m(v - u)$

$\implies \displaystyle \sf \: \Delta p =1( - 8- 8)$

$\implies \displaystyle \sf \: \Delta p =1( - 16)$

$\implies \underline{ \boxed{ \displaystyle \sf \: \Delta p = - 16 \: kg.ms ^{ - 1} }}$

Hence the change in momentum if the ball rebounds with the same speed is -10 kg.m/s.

Now as we know that Force is given by,

$\implies \displaystyle \sf \: F = mass \times Acceleration$

Now as we know that , Acceleration = ( v - u )/t,

$\implies \displaystyle \sf \: F = 1\times \frac{v - u}{t}$

$\implies \displaystyle \sf \: F = \frac{ - 8 - 8}{0.1}$

$\implies \displaystyle \sf \: F = \frac{ - 16}{0.1}$

$\implies \underline{ \boxed{\displaystyle \sf \: F = - 160\: N}}$

Hence The Force applied by the wall is -160 N which means that Force is applied on apposite direction.

Answer 2

Given :-

Mass of ball = 1 kg

Initial Velocity = -8 m/s

Final Velocity = 8 m/s

Time = 0.1 second

To Find :-

Change in momentum

Solution :-

We know that

Change in momentum

$\sf \triangle p = mv - mu$

By taking m as common

$\sf \triangle p = m(v - u)$

Lets put the values

$\sf \triangle p = 1(-8 - 8)$

$\sf \triangle p = (-16)$

Now

We know that

A = v - u/t

a = 8 - (-8)/0.1

a = 16/0.1

a = 160 m/s

Now

F = 1 (160)

F = 160 N

Since it rebound the force will be -160 N