Abowling ball of mass mb = 3.1 kg is rolled down the lane with a velocity of v1 = 2.5 m/s. it strikes a single remaining pin mp = 0.67 kg head on. after the collision the bowling ball has a velocity (in the same direction) of v2 = 1.95 m/s. how fast will the pin move immediately after it is hit?

pin will move in the direction of motion of ball with speed

v = 2.54 m/s

Explanation:

As we know that there is no force on the pin + bowling ball system

So we will have net momentum before and after collision will remain conserved

so we will say that

now plug in all data into the above equation

Here, we are required to determine how fast the pin will move after it is hit.

The pin will move with a velocity of;.

v(p) = 2.54m/s

From the question, it is evident the pin is stationary prior to the collision.

Therefore,

{m(b)×v1} + {m(p)×0} = {m(b)×v2} + {m(p)×v(p)}

We then have;

(3.1 × 2.5) = (3.1 × 1.95) + (0.67 × v(p))

Therefore; 0.67v(p) = 1.705

v(p) = 1.705/0.67

Therefore, the pin will move with a velocity of;

v(p) = 2.54m/s

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distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000