Question

An arrow of 43 g moving at 84 m/s to the right, strikes an apple at rest. The arrow sticks to the apple and both travel at 16.8 m/s after impact. What is the mass of the apple?
(Plz help ASAP!!! I will give you 50 points)

Answer 1

Answer:

The mass of the apple is 0.172 kg (172 g)

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, then the total momentum is the sum of both 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:

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

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

We are given the mass of an arrow m1=43 g = 0.043 kg traveling at v1=84 m/s to the right (positive direction). It strikes an apple of unknown mass m2 originally at rest (v2=0). The common speed after they collide is v'=16.8 m/s.

We need to solve the last equation for m2:

$m_2v_2-m_2v'=m_1v'-m_1v_1$

Factoring m2 and m1:

$m_2(v_2-v')=m_1(v'-v_1)$

Solving:

$\displaystyle m_2=\frac{m_1(v'-v_1)}{v_2-v'}$

Substituting:

$\displaystyle m_2=\frac{0.043(16.8-84)}{0-16.8}$

$\displaystyle m_2=\frac{-2.8896}{-16.8}$

$\displaystyle m_2=0.172\ kg$

The mass of the apple is 0.172 kg (172 g)