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A bowling 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?

Respuesta :

Answer:

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

[tex]m_b v_i = m_b v_{1f} + m_p v_{2f}[/tex]

now plug in all data into the above equation

[tex]3.1(2.5) = 3.1(1.95) + 0.67 v[/tex]

[tex]3.1(2.5 - 1.95) = 0.67 v[/tex]

[tex]v = \frac{3.1(2.5 - 1.95)}{0.67}[/tex]

[tex]v = 2.54 m/s[/tex]

adioabiola

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.

  • During a collision, Momentum is conserved.

Therefore,

  • momentum before collision = momentum after collision.

{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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