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 *?* =
49.0° to the positive *x* axis.

(a) Determine the velocity of the 0.300-kg puck after the collision.

(b) Find the fraction of kinetic energy transferred away or transformed to other forms of energy in the collision.

Answer #1

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 49) + (0.30 kg)(vx) = 0.13 + 0.30 vx

pfy= (0.20 kg)(1.0 m/s)(sin 49) + (0.30 kg)(vy)

= 0.15 + 0.30 vy

Conservation of momentum:

pix = pfx

0.40 kg m/s = 0.13 + 0.30 vx

vx= (0.40 - 0.13)/0.30 = 0.9 m/s

piy = pfy

0 = 0.15 + 0.30 vy

vy= -0.5 m/s

v = [(vx)2 + (vy)2]1/2 = **1.06 m/s**

q = tan-1 (vy /vx ) = -29.05°

(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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