A 0.300-kg puck, initially at rest on a horizontal, frictionless surface, is str
ID: 1537058 • 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 = 56.0° to the positive x axis (see the figure below).
(a) Determine the velocity of the 0.300-kg puck after the collision.
m/s
(b) Find the fraction of kinetic energy transferred away or transformed to other forms of energy in the collision.
Please help, thank you
Before the collision ii After the collision Ulf sin 6 if cos 6 Uof Cos vef sin. p t-- 2fExplanation / 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 56) + (0.30 kg)(vx)
= 0.12 + 0.30 vx
pfy= (0.20 kg)(1.0 m/s)(sin 56) + (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)
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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