Question

Particle 1 of mass 375 g and speed 4.65 m/s undergoes a one-dimensional collision with stationary particle 2 of mass 435 g. What is the magnitude of the impulse on particle 1 if the collision is (a) elastic and (b) completely inelastic?

Answer 1

a) Incase of elastic collision,

P1 = P2

=> M1V1 = M1V1_f + M2V2_f

and 1/2M1V1² = 1/2M1V1_f² + 1/2M2V2_f²

^{from these 2 equations,}

^V1_f = V1 ( M1/M2 - 1)/( M1/M2+1) = 4.65 ( 375/435 - 1) / ( 375/435 +1) = 4.65x ( - 0.138/1.862) = - 0.345 m/s

So P1 = M1V1_f = 0.375x - 0.345 = - 0.13 kg-m/s

So, Impulse = M1x ( V1 - V1_f) = 0.375 x ( 4.65 - (-0.345)) = 0.375 x 4.995 = 1.873 kg-m/s

b) In inelastic collision,

P1 = P2

M1V1 = ( M1 + M2 ) V_f

=> V_f = M1V1/( M1+M2) = 2.18 m/s

Impulse = M1 ( V1 - V_f) = 0.375 x 2.47 = 0.926 kg-m/s