A hockey puck B rests on a smooth surface of ice and is struck by a second puck

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 30.5 from its original direction. (See (Figure 1) .) Puck B acquires a velocity at a 47.0 angle to the original direction of A. The pucks have the same mass.

Part A Compute the speed of puck A after the collision.

Part B Compute the speed of puck B after the collision.

Part C What fraction of the original kinetic energy of puck A dissipates during the collision?

if you can show your work that would be great, just having a hard time following

Explanation / Answer

before collision

initial moemntum L1 = mA*vA1i

after collision

L2 = mA*vA2*cos30.5i + mA*vA2*sin30.5j + mB*vB2*cos47 i - mB*vB2*sin47 j

L2 = L1

mA*vA2*cos30.5i + mA*vA2*sin30.5j + mB*vB2*cos47 i - mB*vB2*sin47 j = mA*vA1 i

mA = mB

comparing j

mA*vA2*sin30.5 + mB*vB2*sin47 = 0

vB2 = vA2*sin30.5/sin47

vB2 = vA2*0.694

comparing i

mA*vA2*cos30.5 + mB*vB2*cos47 = mA*vA1

mA = mB

vA1 = 40 m/s

vA2 *cos30.5 + vB2*cos47 = 40

vA2 *cos30.5 + vA2*0.694*cos47 = 40

vA2 - 30 m/s

vB2 = 30*0.694 = 20.82 m/s

----------------------------

KA1 = (1/2)*mA*vA1^2

kA2 = (1/2)*mA*vA2^2

fraction K2/K1 = vA2^/vA1^2 = (30/40)^2 = 0.5625

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