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

Answer 1

(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.

Answer 2

Let the velocity of the 0.30 kg puck is v m/s at an angle A* to the negative x axis
Thus as the initial y axis momentum = 0
By the law of momentum conservation on y axis for final momentum :-
=>0.2 x sin52* = 0.3 x sinA*
=>sinA* = 0.5253 = sin31.69*
=>A* = 31.69*
On x axis by the law of momentum conservation:-
=>0.2 x 2 = 0.2 x 1 x cos52* + 0.3 x v x cos31.69*
=>v = 1.08512 m/s

2)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.