Question

After being struck by a bowling ball, a 1.3 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.3 kg bowling pin initially at rest. Find the final velocity of the second pin in the following situations:
a) The first pin moves to the right after the collision at 0.8 m/s. Answer in units of m/s.
b) The first pin stops moving when it hits the second pin. Answer in units of m/s.

Answer 1

Answer:

a) 4.2m/s

b) 5.0m/s

Explanation:

This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.

The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses $m_1$ and $m_2$ whose respective velocities before collision are $u_1$ and $u_2$;

$m_1u_1+m_2u_2=m_1v_1+m_2v_2..............(1)$

where $v_1$ and $v_2$ are their respective velocities after collision.

Given;

$m_1=1.3kg\\u_1=5m/s\\m_2=1.3kg\\u_2=0m/s$

Note that $u_2$=0 because the second mass $m_2$ was at rest before the collision.

Also, since the two masses are equal, we can say that $m_1=m_2=m$ so that equation (1) is reduced as follows;

$mu_1+mu_2=mv_1+mv_2\\\\m(u_1+u_2)=m(v_1+v_2)..............(2)$

m cancels out of both sides of equation (2), and we obtain the following;

$u_1+u_2=v_1+v_2.............(3)$

a) When $v_1=0.8m/s$, we obtain the following by equation(3)

$5+0=0.8+v_2\\hence\\v_2=5-0.8\\v_2=4.2m/s$

b) As $m_1$ stops moving $v_1=0$, therefore,

$5+0=0+v_2\\v_2=5m/s$