Marble C, mass 5.0 g, moves at a speed of 20.5 cm/s. It collides with a second m

Question

Marble C, mass 5.0 g, moves at a speed of 20.5 cm/s. It collides with a second marble, D, with mass 10.0 g, moving at 8.0 cm/s in the same direction. After the collision, marble C continues with a speed of 8.0 cm/s in the same direction. Calculate the marbles' momenta before the collision. kg middot m/s (marble C) kg middot m/s (marble D) Calculate the momentum of marble C after the collision. kg middot m/s Calculate the momentum of marble D after the collision. kg middot m/s What is the speed of marble D after the collision? cm/s

Explanation / Answer

Let:
m1 be the mass of marble C,
u1 be its velocity before collision,
v1 be its velocity after collision,
m2 be the mass of marble D,
u2 be its velocity before collision,
v2 be its velocity after collision.

(b)
Momentum of C before collision is:
m1 u1 = 5.0 * 20.5
= 102.5 gm cm/s.

Momentum of D before collision is:
m2 u2 = 10.0 * 8
= 80 gm cm/s.

The total momentum before collision is:
m1 u1 + m2 u2 = 182.5 gm cm/s.

(c)
Momentum of C after collision is:
m1 v1 = 5.0 * 8.0
= 40.0 gm cm/s.

(d)
Conserving momentum:
m1 u1 + m2 u2 = m1 v1 + m2 v2

Momentum of D after collision is:
m2 v2 = m1 u1 + m2 u2 - m1 v1
= 182.5.0 - 40.0
= 142.5 gm cm/s.

(e)
Speed of D after collision is:
v2 = 142.5 / 8.0
= 17.81 cm/s.

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