A hockey puck B rests on a smooth surface of ice and is struck by a second puck
ID: 1598144 • Letter: A
Question
A hockey puck B rests on a smooth surface of ice and is struck by a second puck A, which was originally traveling at 40.0 m/s and which is deflected 28.0 compositefunction from its original direction. (See (Figure 1) Puck B acquires a velocity at a 47.0 ? compositefunction angle to the original direction of A. The pucks have the same mass. Compute the speed of puck A after the collision Express your answer in meters per second to three significant figures. Compute the speed of puck B after the collision Express your answer in meters per second to three significant figures.Explanation / Answer
Given that both A and B has same masses let it be m
initial velocity of A = 40 m / s
initial velocity of B = 0 m /s
total momentum before collision = ( m * 40 ) + ( m * 0 )
= 40 m - ------- ( 1 )
after collision :
velocity of A along horizontal direction = u*cos28 = 0.883 u
velocity of A along vertical direction = u*sin 28 = 0.47 u
velocity of B along horizontal direction = v*cos 47 = 0.682 v
velocity of B along vertical direction = - v sin 47 = - 0.731 v since it is in - ve axis
total momentum along vertical direction = 0
So , m ( 0.47 u ) + m ( - 0.731 v ) =0
0.47 u = 0.731 v
u = 1.556 v ------------( 2 )
total momentum along horizontal direction = m ( 0.883 u ) + m( 0.682 v )
=0.883 * m * 1.556 v + 0.682 m v
= 1.374 m v + 0.682 mv
=2.056 mv ------------------( 3 )
momentum is conserved So, eq ( 2 ) = eq ( 3 )
40 m = 2.056 mv
v = 40 / 2.056
velocity of B = 19.5 m / s
u = 1.556 * v since from ( 2)
= 1.556 * 19.5.
velocity of A = 30.3 m /s
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