Question

8. A ball of mass 0.200 kg has a velocity of 150m/s; a ball of mass 0.300 kg has a

velocity of -0.4m/s. They meet in a head-on elastic collision. (a) Find their velocities

after the collision. (b) Find the velocity of their center of mass before and after the

collision.

Answer 1

(a) Let's apply the law of conservation of momentum:

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

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 0.2 kg ball, $v_1$:

$$v_1=\dfrac{(m_1-m_2)}{(m_1+m_2)}u_1+\dfrac{2m_2}{(m_1+m_2)}u_2,$$ $$v_1=\dfrac{(0.2\ kg-0.3\ kg)}{(0.2\ kg+0.3\ kg)}\cdot1.5\ \dfrac{m}{s}+\dfrac{2\cdot0.3\ kg}{(0.2\ kg+0.3\ kg)}\cdot(-0.4\ \dfrac{m}{s}),$$ $$v_1=-0.78\ \dfrac{m}{s}.$$

The sign minus means that the 0.2 kg ball moves in the opposite direction after the collision (leftward).

Substituting $v_1$into the equation (6) we can find the final velocity of 0.3 kg ball, $v_2$:

$$v_2=\dfrac{2m_1}{(m_1+m_2)}u_1+\dfrac{(m_2-m_1)}{(m_1+m_2)}u_2,$$ $$v_2=\dfrac{2\cdot0.2\ kg}{(0.2\ kg+0.3\ kg)}\cdot1.5\ \dfrac{m}{s}+\dfrac{(0.3\ kg-0.2\ kg)}{(0.2\ kg+0.3\ kg)}\cdot(-0.4\ \dfrac{m}{s}),$$ $$v_2=1.12\ \dfrac{m}{s}.$$

The sign plus also means that the 0.3 kg ball moves in the opposite direction after the collision (rightward).

(b) Let's find the velocity of the center of mass before and after the collision:

$$v_{CM,i}=\dfrac{m_1u_1+m_2u_2}{m_1+m_2},$$ $$v_{CM,i}=\dfrac{0.2\ kg\cdot1.5\ \dfrac{m}{s}+0.3\ kg\cdot(-0.4\ \dfrac{m}{s})}{0.2\ kg+0.3\ kg}=0.36\ \dfrac{m}{s},$$ $$v_{CM,f}=\dfrac{m_1v_1+m_2v_2}{m_1+m_2},$$ $$v_{CM,f}=\dfrac{0.2\ kg\cdot(-0.78\ \dfrac{m}{s})+0.3\ kg\cdot(1.12\ \dfrac{m}{s})}{0.2\ kg+0.3\ kg}=0.36\ \dfrac{m}{s}.$$