Question

1. After being struck by a bowling ball, a 1.5 kg bowling pin moving to the right at +3.0 m/s collides with

Answer 1

Answer:

a) Final velocity of second bowling pin is 2.5m/s.

b) Final velocity of second bowling pin is 3 m/s.

Explanation:

Let 'm' be the mass of both the bowling pin -

m = 1.5 kg

Initial velocity of first bowling pin -

$v_{1} = 3 m/s$

In any type of collision between two bodies in horizontal plane , momentum is conserved along the line of impact.

a) Since , initial velocity of second bowling pin is 0 m/s -

Initial momentum ,

$p_{1} = mv_{1}$

Final velocity of first bowling pin , $v_{2} = 0.5m/s$ [Considering initial direction of motion of the first bowling pin to be positive]

Let $u_{2}$ be the final velocity of the second bowling pin.

∴ Final momentum ,

$p_{2} = mv_{2} + mu_{2}$.

Now ,

$p_{1} = p_{2}$

∴$mv_{1} = mv_{2} + mu_{2}$

∴$u_{2}$ = 3 - 0.5 = 2.5 m/s

∴ Final velocity of second bowling pin is 2.5 m/s.

b) Since , initial velocity of second bowling pin is 0 m/s -

Initial momentum ,

$p_{1} = mv_{1}$

Final velocity of first bowling pin , $v_{2} = 0m/s$ [given][Considering initial direction of motion of the first bowling pin to be positive]

Let $u_{2}$ be the final velocity of the second bowling pin.

∴ Final momentum ,

$p_{2} = mv_{2} + mu_{2}$.

Now ,

$p_{1} = p_{2}$

∴$mv_{1} = mv_{2} + mu_{2}$

∴$u_{2}$ = 3 - 0 = 3 m/s

∴ Final velocity of second bowling pin is 3 m/s.