Question

A ball of 300g is rolled on the ground wit 0.1 N force . it the time taken by the ball to come to rest is 9 seconds then what is the initial velocity of the ball .a) 2m/s b) 5m/s c)3m/s d)4m/s​

Answer 1

Explanation:

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

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