Question

A force acts on a ball for 10 s .and changes its momentum from 12 kg.m/s to 25 Kg.m/s to the positive direction .What is the magnitude of the force

Answer 1

please mark as a BRAINLIST

Let their velocities after the collision be v

1

and v

2

. As we know for elastic collision.

Relative velocity of approach = relative velocity of separation

10−4=v

2

−v

1

⇒6=v

2

−v

1

⇒v

1

=v

2

−6

Applying conservation of momentum,

10×10+5×4=10v

1

+5v

2

120=10v

1

+5v

2

120=10(v

2

−6)+5v

2

=15v

2

−60

15v

2

=180⇒v

2

=12cm/sec

v

1

=6cm/sec