Question

Two carts are on an air-track where friction is negligible. The Incident Cart is moving at an initial velocity of 0.25080m/s, the target cart is at rest. The Incident cart has a mass of 995.8 g and the target cart has a mass of 490.3 g. The carts stick together, the final velocity after collision is 0.077302 m/s. The collision is considered inelastic.

a) What is the Initial Momentum of the carts?

b) What is the Final Momentum of the carts?

c) What is the Initial Kinetic Energy of the carts?

d) What is the Final Kinetic Energy of the carts?

e) What is the % loss of Kinetic Energy?

Please show the formula used.

Answer 1

here,

the mass of 1 , m1 = 995.8 g = 0.9958 kg

the mass of 2 , m2 = 490.3 g = 0.4903 kg

the initial speed of 1 , u1 = 0.2508 m/s

final velocity after the collison , v = 0.077302 m/s

a)

the Initial Momentum of the carts , Pi = m1 * u1

Pi = 0.9958 * 0.2508 kg.m/s = 0.25 kg.m/s

b)

the final Momnetum of the carts , Pf = (m1 + m2) * v

Pf = ( 0.9958 + 0.4903) * 0.077302 = 0.115 kg.m/s

c)

the initial kinetic energy ,KEi = 0.5 * m1 * u1^2

KEi = 0.5 * 0.9958 * 0.2508^2 J = 0.031 kg.m/s

d)

the final kinetic energy , KEf = 0.5 * ( m1 + m2) * v^2

KEf = 0.5 * ( 0.9958 + 0.4903) * 0.077302^2 J

KEf = 3.96 * 10^-3 J

e)

the kinetic energy lost , dKE = KEi - KEf

dKE = 0.027 J

answered by: anonymous