Question

A 2.0 g particle moving at 5.2 m/s makes a perfectly elastic head-on collision with a resting 1.0 g object.
(a) Find the speed of each particle after the collision.(b) Find the speed of each particle after the collision if the stationary particle has a mass of 10 g.(c) Find the final kinetic energy of the incident 2.0 g particle in the situations described in parts (a) and (b).KE in part (a) JKE in part (b) J

Answer 1

Answer:

(a) The speed of the first particle is 1.75 m/s. The speed of the second particle is 6.9 m/s after the collision.

(b) The speed of the first particle is 3.45 m/s in the negative direction. The speed of the second particle is 1.73 m/s.

(c) The final kinetic energy of the incident particle in part (a) and part(b) is 0.0031 J and 0.011 J, respectively.

Explanation:

(a)

In an elastic collision, both momentum and energy is conserved.

$\vec{P}_{initial} = \vec{P}_{final}\\m_1v_1 = m_1v_1' + m_2v_2'\\K_{initial} = K_{final}\\\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2$

Combining these equations will give the speed of the second particle.

$v_2' = \frac{2m_1}{m_1 + m_2}v_1 = \frac{2*2}{2+1}(5.2) = 6.9~m/s$

We can use this to find the speed of the first particle.

$m_1v_1 = m_1v_1' + m_2v_2'\\2(5.2) = 2v_1' + (1)(6.9)\\v_1' = 1.75~m/s$

(b)

If m_2 = 10g.

$v_2' = \frac{2m_1}{m_1 + m_2}v_1 = \frac{2*2}{2+10}(5.2) = 1.73~m/s$

$m_1v_1 = m_1v_1' + m_2v_2'\\2(5.2) = 2v_1' + (10)(1.73)\\v_1' = -3.45~m/s$

The minus sign indicates that the first particle turns back after the collision.

(c)

The final kinetic energy of the particle in part (a) and part (b) is

$K_a = \frac{1}{2}m_1v_1'^2 = \frac{1}{2}(2\times10^{-3})(1.75)^2 = 0.0031 ~J\\K_b = \frac{1}{2}m_2v_1'^2 = \frac{1}{2}(2\times10^{-3})(3.45)^2 = 0.011~J$