Question

A 0.320 kg puck at rest on a horizontal frictionless surface is struck by a 0.220 kg puck moving in the positive x direction with a speed of 4.05 m/s. After the collision, the 0.220 kg puck has a speed of 1.29 m/s at an angle of ? = 60.0° counterclockwise from the positive x axis.

(a) Determine the velocity of the 0.320 kg puck after the collision. Express your answer in vector form.

vf = ___m/s

(b) Find the percent of kinetic energy lost in the collision.

100*?K/Ki = ___%

Answer 1

m1 = 0.32 kg                 m2 = 0.22 kg

before collision

v1i = 0 m/s               v2i = 4.05 m/s

initial linear momentum

Li = m1*v1i + m2*v2i

after collision

V1f = ?                   v2f = 1.29*cos60 i + 1.29*sin60 j

final linear momentum Lf = m1*v1f + m2*v2f

from conservation of linear momentum

Lf = Li

0.22*4.05 i = 0.32*v1f + 0.22*1.29*cos60 i + 0.22*1.29*sin60 j

0.891i = 0.32*v1f + 0.142 i + 0.246 j

v1f = 2.34 i - 0.769 j

=======================

(b)

speed v1f = sqrt(2.34^2+0.769^2) = 2.46 m/s

initial kinetic energy Ki = (1/2)*m1*v1i^2 + (1/2)*m2*v2i^2 = (1/2)*0.22*4.05^2 = 1.8 J

final kinetic energy Kf = (1/2)*m1*v1f^2 + (1/2)*m2*v2f^2

Kf = (1/2)*0.32*2.46^2 + (1/2)*0.22*1.29^2 = 1.15 J

percent of kinetic energy lost in the collision = (Ki - Kf)/ki

percent of kinetic energy lost in the collision. = (1.8 - 1.15)/1.8 *100 = 36.1 %