An object of mass 0.4kg moving with a velocity of 4m/s collides with another object of mass 0.6kg moving in same direction with a velocity of 2m/s. If the collision is perfectly inelastic, what is the loss of K.E. due to impact?
CLASS - XI PHYSICS (Work, Energy and Power)
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m1 = 0.4kg, v1 = 4m/s
m2 = 0.6kg, v2 = 2m/s
Since collision is perfectly inelastic, both masses would move together at velocity v.
According to conservation of momentum,
m1v1+m2v2=(m1+m2)v
⇒0.4×4 + 0.6×2 = (0.6+0.4)v
⇒1.6 + 1.2 = v
⇒ v = 2.8 m/s
initial KE = 0.5m1(v1)² + 0.5 m2(v2)² = 0.5×0.4×4² + 0.5×0.6×2² = 4.4 J
final KE = 0.5(m1+m2)v² = 0.5×1×2.8² = 3.92 J
Loss = initial KE - final KE = 4.4-3.92 = 0.48J
m2 = 0.6kg, v2 = 2m/s
Since collision is perfectly inelastic, both masses would move together at velocity v.
According to conservation of momentum,
m1v1+m2v2=(m1+m2)v
⇒0.4×4 + 0.6×2 = (0.6+0.4)v
⇒1.6 + 1.2 = v
⇒ v = 2.8 m/s
initial KE = 0.5m1(v1)² + 0.5 m2(v2)² = 0.5×0.4×4² + 0.5×0.6×2² = 4.4 J
final KE = 0.5(m1+m2)v² = 0.5×1×2.8² = 3.92 J
Loss = initial KE - final KE = 4.4-3.92 = 0.48J
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