Question

Problem 3 (25 pts) A hockey puck B (400.0 g) is initially at rest on a frictionless, level ice and is struck by a second puck A (300.0 g), which was originally traveling at 8.00 m/s. Puck B acquires a velocity of 4.00 m/s at a 40.0° angle to the original velocity of A as in the diagram. Compute the velocity, magnitude and direction, of A after the collision. 8 m/s A A 40 4.00 m/s

Answer 1

apply linear momentum conservation in both directions

suppose puck A make angle theta wiht horizontal line and moving with speed V

in X

Pi = Pf  

Ma* Va = Ma*Vaf + Mb*Vbf

0.300*8 = 0.300*V*cosθ + 0.400*4*cos40

0.300*V*cosθ = 0.300*8 - 0.400*4*cos40  = 1.17433

V*cosθ = 1.17433/0.3 =3.914433 .......i

in Y

Pi = Pf  

Ma* Va = Ma*Vaf + Mb*Vbf

0 = 0.300*V*sinθ - 0.400*4*sin40

V*sinθ = 0.400*4*sin40 /0.300 =3.4282 ......ii

from eq. i and ii

V*sinθ/V*cosθ = 3.4282/3.914433

tanθ = 3.4282/3.914433

θ = tan^-1(3.4282/3.914433) =41.21°  

so direction =41.21° from horizontal counterclockwise

V*sinθ =3.4282

V = 3.4282/sin41.21 =5.20354 m/s

V = 5.20354 m/s ansswer

****************************************************************************
Goodluck for exam Comment in case any doubt, will reply for sure..