Question

A ball of mass 1 = 0.250 kg and velocity 1 = 5.00 m/s [E] collides head-on with a ball of mass 2 = 0.800 kg that is initially at rest. No external forces act on the balls. a) Show what is conserved through the appropriate formula if the collision is elastic. b) What are the velocities of the balls after the collision?

a) a) Momentum, b) 1 = 1.25 m/s [W], 2 = 1.25 m/s [E]
b) a) Energy, b) 1 = 2.50 m/s [W], 2 = 0 m/s
c) a) Angular momentum, b) 1 = 5.00 m/s [E], 2 = 0 m/s
d) a) Kinetic energy, b) 1 = 0 m/s, 2 = 5.00 m/s [W]

Answer 1

In an elastic collision, momentum is conserved. The velocities of the balls after the collision are 1.25 m/s [W] and 1.25 m/s [E].

Step-by-step explanation:

a) In an elastic collision, momentum is conserved. The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be expressed as:

m1*v1 + m2*v2 = m1*v1' + m2*v2'

b) To determine the velocities of the balls after the collision, we can use the conservation of momentum equation and the given information. Plugging in the values:

(0.250 kg)*(5.00 m/s) + (0.800 kg)*(0 m/s) = (0.250 kg)*(v1') + (0.800 kg)*(v2')

Solving for v1' and v2', we find that v1' = 1.25 m/s [W] and v2' = 1.25 m/s [E].