Question

Before After y 相A hockey puck of mass m = 0.170 kg is loaded into a spring gun with spring constant k = 306 N/m. The spring is compressed by a distance d = 0.100 m and then released, launching the puck onto a horizontal and frictionless surface of ice with speed v in the positive x-direction. This puck then collides with another puck of the same mass which is at rest at the origin.

After the collision the two pucks move away from the origin with the same speed v' at angles +? and −?, respectively, as measured from the positive x-axis as shown.

1. What was v, the speed of the first puck before the collision?

2. If the collision was elastic, what is v', the speed of the two pucks after the collision?

3. If the collision was elastic, what is ??

4. If ? = 30º, what fraction of the initial kinetic energy was lost in the collision?

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Answer #1

1.

v = speed of first puck before collision

d = compression of the spring = 0.100 m

k = spring constant = 306 N/m

m = mass of puck = 0.170 kg

using conservation of energy

kinetic energy gained by the puck = spring potential energy stored in spring when compressed

(0.5) m v2 = (0.5) k d2

m v2 = k d2

(0.170) v2 = (306) (0.1)2

v = 4.24 m/s

2)

using conservation of kinetic energy

(0.5) m v2 = (0.5) m v'2 + (0.5) m v'2

(0.5) v2 = v'2

(0.5) (4.24)2 = v'2

v' = 3 m/s

3)

using conservation of momentum along the X-direction

m v = m v' Cos\theta + m v' Cos\theta

m v = 2 m v' Cos\theta

v = 2 v' Cos\theta

4.24 = 2 (3) Cos\theta

\theta = 45 deg

4)

v'' = final speed when angle is 30

using conservation of momentum along the X-direction

m v = m v'' Cos\theta + m v'' Cos\theta

m v = 2 m v'' Cos\theta

v = 2 v'' Cos\theta

4.24 = 2 v'' Cos30

v'' = 2.45 m/s

fraction of energy lost is given as

fraction = (Final kinetic energy - initial kinetic energy)/initial kinetic energy

fraction = (((0.5) m v''2 + (0.5) m v''2) - (0.5) m v2) /((0.5) m v2)

fraction = (2 v''2 - v2) /( v2)

fraction = (2 (2.45)2 - (4.24)2) /( (4.24)2)

fraction = - 0.33

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