Question

In a particle accelerator, two particles with masses such that m₁ = 3m₂ collide. Particle 1 is initially moving at 1.5 x 10⁷ ms^-1 and particle 2 is moving in the same direction at 0.5 x 10⁷ ms^-1

a. Calculate the final speed of each particle assuming the collision is elastic.

b. Calculate the final speed of each particle assuming the collision is perfectly inelastic.

Answer 1

(a)  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 $v_1$:

$$v_1=\dfrac{(m_1-m_2)}{(m_1+m_2)}u_1+\dfrac{2m_2}{(m_1+m_2)}u_2,$$ $$v_1=\dfrac{(3m_2-m_2)}{(3m_2+m_2)}\cdot1.5\cdot10^7\ \dfrac{m}{s}+\dfrac{2m_2}{(3m_2+m_2)}\cdot0.5\cdot10^7\ \dfrac{m}{s},$$ $$v_1=1.0\cdot10^7\ \dfrac{m}{s}.$$

Substituting $v_1$into the equation (6) we can find $v_2$:

$$v_2=\dfrac{2m_1}{(m_1+m_2)}u_1+\dfrac{(m_2-m_1)}{(m_1+m_2)}u_2,$$ $$v_2=\dfrac{2\cdot3m_2}{(3m_2+m_2)}\cdot1.5\cdot10^7\ \dfrac{m}{s}+\dfrac{(m_2-3m_2)}{(3m_2+m_2)}\cdot0.5\cdot10^7\ \dfrac{m}{s},$$ $$v_2=2.0\cdot10^7\ \dfrac{m}{s}.$$

b) From the law of conservation of momentum, we have:

$$m_1u_1+m_2u_2=(m_1+m_2)v_f,$$ $$v_f=\dfrac{m_1u_1+m_2u_2}{m_1+m_2},$$ $$v_f=\dfrac{3m_2u_1+m_2u_2}{3m_2+m_2}=\dfrac{3u_1+u_2}{4},$$ $$v_f=\dfrac{3\cdot1.5\cdot10^7\ \dfrac{m}{s}+1.5\cdot10^7\ \dfrac{m}{s}}{4}=1.5\cdot10^7\ \dfrac{m}{s}.$$