A blue puck with a mass of 4.10×10 2 kg , sliding with a velocity of 0.210 m/s N

Question

A blue puck with a mass of 4.10×102  kg , sliding with a velocity of 0.210  m/s North on a frictionless, horizontal air table, makes a perfectly elastic, head-on collision with a red puck with mass m, initially at rest. After the collision, the velocity of the blue puck is 6.0×102  m/s North.

Part A) Find the momentum of the red puck after the collision. Give the horizontal component with North positive.

Give the horizontal component of momentum with North positive.

Part B) Find the kinetic energy of the red puck after the collision.

Explanation / Answer

ANALYSIS:
1. Momentum is conserved (law of conservation of momentum)
2. Kinetic energy is conserved (since collision is elastic)

SOLUTION:
Using the law of conservation of momentum
M1V1 + M2V2 = M1V3 + M2V4
where
M1 = mass of the blue puck = 0.041 kg.
V1 = initial velocity of the blue puck = 0.210 m/sec.
M2 = mass of the red puck
V2 = initial velocity of the red puck = 0 (at rest)
V3 = final velocity of the blue pack = 0.060 m/sec.
V4 = final velocity of the red puck

Substituting values,
0.041(0.210) + 0 = 0.041(0.06) + (M2)(V4)
(M2)(V4) = 0.00615 --- call this Equation 1

Since the kinetic energy is conserved,
(1/2)(0.041)(0.210)^2 = (1/2)(0.041)(0.06)^2 + (1/2)(M2)(V4)^2
0.0018081 = 0.0001476 + (M2)(V4)^2
(M2)(V4)^2 = 0.0016605

The above can be rewritten as
(M2*V4)(V4) = 0.0016605
Since (M2*V4) = 0.00615 (from Equation 1), then
0.00615(V4) = 0.0016605
and solving for "V4"
V4 = 0.27 m/sec

A) the momentum of the red puck after the collision = m4*V4 = 0.00615 kgm/s

the horizontal component of momentum = 0

B) the kinetic energy of the red puck after the collision = 0.5*m4*V4^2 = 0.5*0.0016605 = 0.00083025 J

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