A cart of mass m 1 = 2.00 kg rolls along the floor with a velocity of 14.0 m/s i

Question

A cart of mass m1 = 2.00 kg rolls along the floor with a velocity of 14.0 m/s in the positive direction. It hits a second cart, of mass 5.0 kg, and bounces back with a velocity of 6.00 m/s in the opposite direction. If the second cart was initially at rest, what is its final velocity? Is the collision elastic or inelastic & why?

A.) 2.00 m/s

B.) 4.00 m/s

C.) 8.00 m/s

For second question:

A.) Elastic, because kinetic energy is conserved.

B.) Inelastic, because momentum is not conserved.

C.) Inelastic, because kinetic energy is not conserved.

Explanation / Answer

given
m1 = 2 kg, m2 = 5 kg
u1 = 14 m/s, u2 = 0
v1 = 6 m/s, v2 = ?
from the conservation of linear momentum
initial momentum = final momentum
Pinitial = Pfinal
(m1 x u1) + (m2 x u2) = (m1 x v1)+(m2 x v2)
(2x14) + (5*0) = (2x(-6)) + (5xv2)
v2 = 8 m/s Option (C)

B) It is elastic collision, because linear momentum and kinetic energy is conserved.

initial KE = (1/2 x m1 x u12) + (1/2 x m2 x u22)
= (0.5x2x142) + (0.5x5x0) = 196 J

final KE = (1/2 x m1 x v12) + (1/2 x m2 x v22)
= (0.5x2x62) + (0.5x5x82) = 196 J

initial momentum = final momentum
initial KE = final KE

So, It is elastic collision

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