Question

A football, 1.25 kg, is thrown at a golf ball, 75 gm. The football is moving at 190.0 m/s to the right while the golf ball is moving 35.0 m/s to the left. What are their velocities after an elastic collision in one dimension?

Answer 1

Let’s find the formula for the final velocities of the football and golf ball in case of elastic collision. From the law of conservation of momentum, we have:

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

Since collision is elastic, kinetic energy is conserved and we can write:

$$\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2. (2)$$

Let’s rearrange equations (1) and (2):

$$m_1(u_1-v_1)=m_2(v_2-u_2), (3)$$ $$m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2). (4)$$

Let’s divide equation (4) by equation (3):

$$\dfrac{(u_1-v_1)(u_1+v_1)}{u_1-v_1}=\dfrac{(v_2-u_2)(v_2+u_2)}{v_2-u_2},$$ $$u_1+v_1=u_2+v_2. (5)$$

Let's express $v_2$ from the equation (5) in terms of $u_1$, $u_2$ and $v_1$:

$$v_2=u_1-u_2+v_1. (6)$$

Let’s substitute equation (6) into equation (3). After simplification, we get:

$$(m_1-m_2)u_1+2m_2u_2=(m_1+m_2)v_1.$$

From this equation we can find the final velocity of the football, $v_1$:

$$v_1=\dfrac{(m_1-m_2)}{(m_1+m_2)}u_1+\dfrac{2m_2}{(m_1+m_2)}u_2,$$ $$v_1=\dfrac{(1.25\ kg-0.075\ kg)}{(1.25\ kg+0.075\ kg)}\cdot190\ \dfrac{m}{s}+\dfrac{2\cdot0.075\ kg}{(1.25\ kg+0.075\ kg)}\cdot(-35\ \dfrac{m}{s}),$$ $$v_1=164.53\ \dfrac{m}{s}.$$

Substituting $v_1$into the equation (6) we can find the final velocity of the golf ball, $v_2$:

$$v_2=\dfrac{2m_1}{(m_1+m_2)}u_1+\dfrac{(m_2-m_1)}{(m_1+m_2)}u_2,$$ $$v_2=\dfrac{2\cdot1.25\ kg}{(1.25\ kg+0.075\ kg)}\cdot190\ \dfrac{m}{s}+\dfrac{(0.075\ kg-1.25\ kg)}{(1.25\ kg+0.075\ kg)}\cdot(-35\ \dfrac{m}{s}),$$ $$v_2=389.53\ \dfrac{m}{s}.$$