Question

A ball of mass 2 kg is dropped from a height of 5 m. On striking the ground it rebounds with same speed

with which it struck the ground. Find the magnitude of net change in its momentum just after striking the

ground​

Answer 1

Answer:

0

Explanation:

intial momentum =m1u1

final momentum=m1v1

given v1=u1

magnitude change in net momentum is m1v1-m1u1

∴ as v1=u1

m1v1-m1v1=0

Answer 2

Given:

A ball of mass 2 kg is dropped from a height of 5 m. On striking the ground it rebounds with same speed with which it struck the ground.

To find:

Magnitude of change in Momentum of ball.

Calculation:

Let the Velocity of the ball just before hitting the ground be v ;

$\rm{ \therefore {v}^{2} = {u}^{2} + 2gh}$

$\rm{ = > {v}^{2} = {(0)}^{2} +( 2 \times 10 \times 5)}$

$\rm{ = > {v}^{2} = 0 +100}$

$\rm{ = > {v}^{2} = 100}$

$\rm{ = > v = \sqrt{100} }$

$\rm{ = > v = 10 \: m {s}^{ - 1} }$

Now , after striking the ground the ball rebounds with same magnitude of Velocity.

So, net change in magnitude of Momentum

$= > |\Delta P| = m(10 - 10) = 0$

So, there is no change in Momentum.