A 0.300-kg puck, initially at rest on a horizontal, frictionless surface, is str
ID: 1290702 • Letter: A
Question
A 0.300-kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.200-kg puck moving initially along the x axis with a speed of 2.00 m/s. After the collision, the 0.200-kg puck has a speed of 1.00 m/s at an angle of ? = 52.0
A 0.300-kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.200-kg puck moving initially along the x axis with a speed of 2.00 m/s. After the collision, the 0.200-kg puck has a speed of 1.00 m/s at an angle of ? = 52.0° to the positive x axis (see the figure below).(a) Determine the velocity of the 0.300-kg puck after the collision. 1 m/s (b) Find the fraction of kinetic energy transferred away or transformed to other forms of energy in the collision.Explanation / Answer
(a) Determine the velocity of the other puck after the collision?
Solution:
pix = (0.20 kg)(2.0 m/s) = 0.40 kg m/s
piy = 0
After collision:
pfx= (0.20 kg)(1.0 m/s)(cos 53) + (0.30 kg)(vx)
= 0.12 + 0.30 vx
pfy= (0.20 kg)(1.0 m/s)(sin 53) + (0.30 kg)(vy)
= 0.16 + 0.30 vy
Conservation of momentum:
pix = pfx
0.40 kg m/s = 0.12 + 0.30 vx
vx= (0.40 - 0.12)/0.30 = 0.93 m/s
piy = pfy
0 = 0.16 + 0.30 vy
vy= -0.53 m/s
v = [(vx)2 + (vy)2]1/2 = 1.07 m/s
q = tan-1 (vy /vx ) = -30.0o
(b) Find the fraction of kinetic energy lost in the collision.
Solution:
Ki = (1/2)mv2 = (0.5)(0.20 kg)(2.0 m/s)2 = 0.40 J
Kf = (1/2)mv12 + (1/2)mv22 = 0.10 J + 0.17 J = 0.27 J
Ki - Kf = 0.13
Fraction = (0.13)/(0.40) = 0.325 lost.
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