A small block of mass 20.0 grams is moving to the right on a horizontal friction

Question

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.580 m/s. The block has a head-on elastic collision with a 40.0 gram block that is initially at rest. Since the collision is head-on, all velocities lie along the same line, both before and after the collision.

(a) What is the speed of the 20.0 gram block after the collision? m/s

(b) After the collision is the 20.0 gram block moving to the left or to the right?

(c) What is the speed of the 40.0 gram block after the collision? m/s

(d) After the collision is the 40.0 gram block moving to the left or to the right?

Explanation / Answer

First, conserve momentum:
20g * 0.7m/s + 0 = 20g * u + 40g * v
14 = 20u + 40v

For an elastic, head-on collision, we know (from conservation of energy), that the
relative velocity of approach = relative velocity of separation, or
0.7 m/s = v - u
v = u + 0.7
plug this into the momentum equation
14 = 20u + 40(u + 0.7) = 60u + 28
u = -14 / 60 = -0.233 m/s

(a)
speed = 0.233 m/s

(b)
The negative sign means "to the left"

(c)
v = u + 0.7 m/s = -0.233 + 0.7 = 0.467 m/s

(d)
to the right

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