Question

A arrow of mass 0.20 kg is shot at 32.2 m/s. It hits an apple (m = 0.77 kg), and goes through the apple, leaving with a velocity of 18.3 m/s.

What is the velocity of the apple after the arrow leaves it?

answer in m/s

Answer 1

Answer:

The apple travels at 3.6 m/s

Explanation:

The Law Of Conservation Of Linear Momentum

The total momentum of a system of bodies is conserved unless an external force is applied to it.

The formula for the momentum of a body with mass m and speed v is:

P = mv

If we have a system of two bodies the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

Solving for v2':

$\displaystyle v'_2=\frac{m_1v_1+m_2v_2-m_1v'_1}{m_2}$

The arrow has a mass of m1=0.2 kg and travels at v1=32.2 m/s. It hits an apple (assumed stationary at v2=0) of mass m2=0.77 kg and continues through the apple with a speed of v1'=18.3 m/s. We'll calculate the speed of the apple after the hit.

$\displaystyle v'_2=\frac{0.2*32.2+0.7*0-0.2*18.3}{0.77}$

$\displaystyle v'_2=\frac{2.78}{0.77}$

$v'_2=3.6 \ m/s$

The apple travels at 3.6 m/s