Question

A ball of mass

0.1kg,

starting from rest, falls a height of 4.0 m and then collides with

the ground. After the collision, the ball bounces up to a height of 2.0 m. The collision

with the ground takes place over a time

4.0 10 s.

3



Determine (i) the momentum of

the ball immediately before the collision and immediately after the collision and (ii) the

average force exerted by the ground on the ball. Take

10.0 ms .

2 g 

​

Answer 1

Answer:

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Answer 2

(I). The momentum before and after the collision is 0.885 kg m/s and 0.626 kg m/s,

(II). The force exerted by the ground on the ball is 15.11 N.

Explanation:

Given that,

Mass of ball = 0.1 kg

Height = 4.0 m

Bounce height = 2.0

Time = 10 ms

(I). We need to calculate the momentum before the collision

Using formula of momentum

$P_{1}=m\sqrt{2gh_{1}}$

Put the value into the formula

$P_{1}=0.1\sqrt{2\times9.8\times4.0}$

$P_{1}= 0.885\ kg m/s$

We need to calculate the momentum after the collision

Using formula of momentum

$P_{2}=m\sqrt{2gh_{2}}$

Put the value into the formula

$P_{2}=0.1\sqrt{2\times9.8\times2.0}$

$P_{2}= 0.626\ kg m/s$

(II). We need to calculate the force exerted by the ground on the ball

Using formula of force

$F=\dfrac{p_{1}+p_{2}}{t}$

Put the value into the formula

$F=\dfrac{0.885+0.626}{0.1}$

$F=15.11\ N$

Hence, (I). The momentum before and after the collision is 0.885 kg m/s and 0.626 kg m/s,

(II). The force exerted by the ground on the ball is 15.11 N.

Learn more :

Topic : momentum

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