Question

A 10 g marble slides to the left at a speed of 0.4 m/s along a frictionless surface. It has a head-on, elastic collision with a larger, 30 g marble sliding to the right at 0.2 m/s. Since the collision is head-on, all motion is along the x-axis.

a. Find the total kinetic energy AND the total momentum before the collision

b. The marbles bounce off each other elastically. Each marble has a different velocity after the collision. The lighter marble ends up with a velocity with a magnitude five times that of the heavier marble. Find the velocity of each marble after the collision.

c. Calculate the change in kinetic energy for each individual marble.

d. Calculate the change in momentum for each individual marble.

Answer 1

Assuming motion along left as negative, motion towards right as positive,

Given, Mass of lighter marble, m1 = 10 g = 0.01 kg

Initial speed of lighter marble , u1 = - 0.4 m/s

Mass of larger marble, m2 = 30 g = 0.03 kg

Initial speed of larger marble , u2 = 0.2 m/s

a)

Total kinetic energy before collision, KE_total = kinetic energy of marble 1 before collision + kinetic energy of marble2 before collision

= 1/2 * m1*u1² + 1/2*m2*u2²

= 1/2 * 0.01 * (-0.4)² + 1/2 *0.03*0.2²

= 1.4 * 10^-3 J

Total momentum before collision, P_total = momentum of marble 1 before collision + momentum of marble 2 before collision

= m1 * u1 + m2*u2

= 0.01 *( - 0.4 ) + 0.03 * 0.2 = 2 * 10^-3 kgm/s

b)

Since the collision is eleatic head-on collision, so after collision velocity of marble 1 will be positive (towards right) and velocity of marble 2 will be towards left so negative

let V1 and V2 be the velocity of marble1 and marble 2 after collision,

as given, V1 = 5* V2

apply the law of conservatin of momentum just before collision and just after collision,

Momentum of the system just before collision = Momentum of the system after the collision

2 * 10^-3 = 0.01 * V1 + 0.03 * (- V2)

2 * 10^-3 = 0.01 * 5*V2 - 0.03 *V2

so, velocity of heaviermarble2, V2 = 0.1 m/s towards left

so, velocity of lighter marble1, V1 = 5 * 0.1 = 0.5 m/s towards right

c)

Change in kinetic energy for marble1,

KE1 = 1/2 * m1 * (v1² - u1²)

= 1/2 * 0.01 * (0.5² - (-0.4)²) = 4.5 * 10^-4 J

Change in kinetic energy for marble2,

KE2 = 1/2 * m2 * (v2² - u2²)

= 1/2 * 0.03 * ((-0.1)² - 0.2²) = - 4.5 * 10^-4 J

d)

Change in momentum for marble1,

P1 = m1 * (v1 - u1)

= 0.01 * (0.5 - (-0.4)) = 9 * 10^-3 J

Change in momentum for marble2,

P2 = m2 * (v2 - u2)

= 0.03 * (-0.1- 0.2) = - 9 * 10^-3 J