A moving mass of 8 kg collides elastically with a stationary mass of 2

1.	A moving mass of 8 kg collides elastically with a stationary mass of 2 kg. If E be the initial kineticenergy of the mass, the kinetic energy left with it after collision will be(a) 0.80 E(b) 0.64 E(c) 0.36 E(d) 0.08 E
Answer» answer : option (c) 0.36E explanation : Let v1' is final speed of mass 8kg and v2' is final speed of mass 2kg. INITIALLY, mass of 2kg is in rest so, v2 = 0 Initial kinetic energy = 1/2 (8kg)v1² or, E = 4v1² or, v1 = √E/2 using formula, v2' = 2m1/(M1 + m2)v1 - (m1 - m2)/(m1 + m2)v2 = 2 × 8 /(8 + 2) × √E/2 = 8√E/10 v1' = (m1 - m2)/(m1 + m2) v1 + 2m2/(m1 + m2) v2 = (8 - 2)/(8 + 2) × √E/2 + 0 = 3√E/10 now, final kinetic energy of mass 8kg, E' = 1/2 (8kg) × (3√E/10)² = 4 × 9E/100 = 36E/100 = 0.36 E hence, kinetic energy left with it after collision is 0.36E

A moving mass of 8 kg collides elastically with a stationary mass of 2 kg. If E be the initial kineticenergy of the mass, the kinetic energy left with it after collision will be(a) 0.80 E(b) 0.64 E(c) 0.36 E(d) 0.08 E

Answer»

answer : option (c) 0.36E

explanation : Let v1' is final speed of mass 8kg and v2' is final speed of mass 2kg.

INITIALLY, mass of 2kg is in rest so, v2 = 0

Initial kinetic energy = 1/2 (8kg)v1²

or, E = 4v1²

or, v1 = √E/2

using formula,

v2' = 2m1/(M1 + m2)v1 - (m1 - m2)/(m1 + m2)v2

= 2 × 8 /(8 + 2) × √E/2

= 8√E/10

v1' = (m1 - m2)/(m1 + m2) v1 + 2m2/(m1 + m2) v2

= (8 - 2)/(8 + 2) × √E/2 + 0

= 3√E/10

now, final kinetic energy of mass 8kg, E' = 1/2 (8kg) × (3√E/10)²

= 4 × 9E/100

= 36E/100

= 0.36 E

hence, kinetic energy left with it after collision is 0.36E

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