A hockey puck has a velocity of v = v x i + v y j . Another puck with identical

Question

A hockey puck has a velocity of v = vxi + vyj. Another puck with identical mass is at rest at the origin. The two pucks collide. After the collision it is observed that the second puck has a velocity of vf = v2yj.

Randomized Variablesvx = 4.6 m/s
vy = 4.2 m/s
v2y = 1 m/s (a) Write an expression for the first puck's velocity after the collision, in terms of the variables in the problem statement and the unit vectors i and j.
(b) What is the magnitude of the first puck's velocity, in meters per second, after the collision? (c) At what angle does this velocity make with respect to the x-axis, in degrees?

Explanation / Answer

as the external force on the system is zero so the momentum of the system will remain conserve, if each block has mass 'm' then -

mu + mv = mu' + mv'

u + v = u' + v'

here u = 4.6 i + 4.2 j v = 0

v' = 1 j

4.6 i + 4.2 j = u' + j

u' = 4.6 i + 3.2 j

magnitude of u' = 5.60 m/sec

angle from x axis, theta = tan-1(3.2/4.6)

theta = 34.82 degrees

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