Question

A ball of mass 20 g moving at 10 m s¹ horizontally collides with another ball of mass 40 g that is at rest. After collision the 20 g ball has a velocity of 4 m s at an angle of 50° relative to its original direction. Calculate the total change in kinetic energy of the system. A. 1.21 x 10-21 B. -121 x 10-21 C. 112 x 10-2 D. 815 x 102 E. -518 x 10-1​

Answer 1

We can use conservation of momentum and conservation of energy to solve this problem.

Conservation of momentum:

Initial momentum = final momentum

m1v1 + m2v2 = (m1 + m2)v'

where m1 = 20 g = 0.02 kg, v1 = 10 m/s, m2 = 40 g = 0.04 kg, v2 = 0, v' is the final velocity of both balls after collision.

Solving for v':

v' = (m1v1 + m2v2)/(m1 + m2)

v' = (0.02 x 10 + 0.04 x 0)/(0.02 + 0.04)

v' = 6.67 m/s

Conservation of energy:

Initial kinetic energy = final kinetic energy

(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)(m1 + m2)v'^2

where v' is the final velocity calculated from conservation of momentum.

Solving for the total change in kinetic energy:

ΔKE = (1/2)(m1v1^2 + m2v2^2) - (1/2)(m1 + m2)v'^2

ΔKE = (1/2)(0.02 x 10^2 + 0.04 x 0^2) - (1/2)(0.02 + 0.04) x 6.67^2

ΔKE = -0.518 J

Therefore, the answer is (E) -518 x 10^-1.