Question

4) Block 1 of mass 200 kg slides over a frictionless surface with a velocity of 0.25 cm/s and strikes block 2 of mass 100 kg sliding to the left at 0.75 cm/s. What is the final velocity of each block if the collision is

A) Perfectly elastic?

B) Perfectly inelastic?

C) If there is an external force of 4000 kg*cm/s^2 to the right for 0.008 s during the perfectly inelastic collision?

Answer 1

4)
m1 = 200 kg
u1 = 0.25 cm/s
m2 = 100 kg
u2 = -0.75 cm/s

a) let v1 and v2 are the velocities of v1 and v2 after the perfectly elastic collision.

v1 = ( (m1 - m2)*u1 + 2*m2*u2 )/(m1 + m2)

= ( (200 - 100)*0.25 + 2*100*(-0.75) )/(200 + 100)

= -0.42 cm/s <<<<<<<<--------------Answer

v2 = ( (m2 - m1)*u2 + 2*m1*u1 )/(m1 + m2)

= ( (100 - 200)*(-0.75) + 2*200*0.25 )/(200 + 100)

= 0.58 cm/s <<<<<<<<--------------Answer

b) let v is the final velocity of both objects in perfectly inelastic collision.

Apply conservation of momentum

m1*u1 + m2*u2 = (m1 + m2)*v

v = (m1*u1 + m2*u2)/(m1 + m2)

= (200*0.25 + 100*(-0.75))/(200 + 100)

= -0.083 m/s <<<<<<<<--------------Answer

c) now Apply Impulse-momentum theorem

Impulse = change in momentum

F*t = (m1 + m2)*(vf - vi)

4000*0.008 = (200 + 100)*(v - (-0.083))

==> v = 0.024 cm/s <<<<<<<<--------------Answer