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?
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 + m v'
Cos
m v = 2 m v' Cos
v = 2 v' Cos
4.24 = 2 (3) Cos
= 45 deg
4)
v'' = final speed when angle is 30
using conservation of momentum along the X-direction
m v = m v'' Cos + m v''
Cos
m v = 2 m v'' Cos
v = 2 v'' Cos
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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