Question

2.0-g particle moving at 5.8 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.

Answer 1

Explanation:

Apply the law of conservation of momentum as shown:

$m_1u_1 +m_2u_2 = (m_1+m_2)v$ where:

m1 is the mass of the particle = 2.0g

u1 is the velocity of the particle = 5.8m/s

m2 is the mass of the object = 1.0g

u2 is the velocity of the resting object = 0m/s

v is their common velocity after collision = ?

Substitute

2(5.8)+1(0) = (2+1)v

11.6 + 0 = 3v

11.6 = 3v

v = 11.6/3

v = 3.87m/s

Hence the speed of each particle after the collision is 3.87m/s