Question

A force is exerted on a body of mass 10 kg, initially at rest. If its velocity changes to 5m/s in a short interval of 10seconds, find (i) initial momentum (ii) change in momentum (iii) force exerted
pls help 26 points

Answer 1

Answer:

For a perfectly elastic collision, coefficient of restitution is e=1.

Let u and v be the velocities of 10 kg and 1 g masses after collision.

Thus, e=

rel. vel. of approach

rel. vel.of separation

=

5−0

v−u

=

5

v−u

But, as e=1, we have v−u=5.

Total momentum before collision is P=10×5=50 kg m/s

Total momentum after collision is P=10×(v−5)+0.001×v=10.001v−50

Thus, 10.001v−50=50⇒v=

10.001

100

≈10 m/s