Question

A iron ball of mass 50g falls from a height of 50m and rises upto height 3.2m after colloiding with the horizontal surface . If the point of contact is 0.03 second. Find the change in momentum.

Answer 1

Answer:

Explanation:

Given that,

Ball of mass = 50 g

Height = 1 m

Rebound to height = 0.5 m

Time = 0.1 s

Now,

Initial velocity of the ball =um/s

Let the final velocity =vm/s

Using equation of motion

v

2

=u

2

+2gh

v

2

=0+2×9.8×0.1

v=

1.96

​

v=1.4m/s

In the second time,

Final velocity=0m/s

Now, Using equation of motion

v

2

=u

2

−2gh

0=u

2

−2×9.8×0.5

u

2

=9.8

u=3.1m/s

Now, Impulse= change in linear momentum

=mv−m(−u)

=m(v+u)

=0.05×(1.4+3.1)

=0.225Ns

Now, the Force=Impulse/Time

F=

time

impulse

​

F=

0.1

0.225

​

F=2.25N

Hence, the impulse is 0.225 Ns and force is 2.25 N