Question

A 2 kg puck, initaly at rest on a horizonta, frictionless surface is struck by another 2 kg puck moving at a velocity of 2 m/s along the x-axis. After the collision, one puck is moving at a speed of 1 m/s at an angle 53 degrees to the positive x-axis.

A. What is the velocity of the other puck?

B. How much energy is lost in the collision?

C. At what angle, does the other puck make?

Answer 1

oblique

m1 (puck 1) = 2 kg                m2(puck 2 ) = 2 kg

before collision

speeds

u1x = 0 m/s                        u2x = 2 m/s

u1y = 0                             u2y = 0

after collision

v1x = v1*costheta = 1*cos53                     v2x = ?

v1y = v1*sintheta = 1*sin53                    v2y = ?

from momentum conservation

along y momentum is conserved

momentum beifre collision = momentum after collision

Piy = Pfy

m1*u1y + m2*u2y = m1*v1y + m2*v2y

0 = m*v1y + m*v2y

-1*sin53 = v2y

v2y = -0.798 m/s

along x axis momentum is conserved

momentum before collision = momentum after collision

Pix = Pfx

m1*u1x + m2*u2x = m1*v1x + m2*v2x

m*0 + m*2 = m*cos53 + m*v2x

2 = cos53 + v2x

v2x = 1.4 m/s

A)

velocity of other puck = sqrt(0.798^2+1.4^2) = 1.61 m/s

B)

energy lost = ((1/2)*m1*v1^2 + (2/2)*m2*v2^2 ) - (1/2)*m1*u1^2

energy lost = (1/2)*2*1^2 + (1/2)*2*1.61^2 - (1/2)*2*2^2 = -0.408 J

energy lost = 0.408 J

(C)

angle = tan^-(V2y/v2x) = 29.5 degrees

DONE please check the answer. any doubts post in comment box