Question

A hockey puck B rests on frictionless, level ice and is
struck by a second puck A, which was originally traveling at
40.0 m/s and which is deflected 30.0° from its original
direction. (See the Figure). Puck B acquires a velocity at a
45.0° angle to the original direction of A. The pucks have the
same mass. (a) Compute the speed of each puck after the
collision. (b) What fraction of the original kinetic energy of
puck A dissipates during the collision?

Answer 1

Let Mass of each block = m kg

final velocity of Puck A = VA m/s

final velocity of Puck B = VB m/s

By using horizontal momentum conservation

momentum before collision = mumentum after collision

muA + muB  = mVA cos30o +VB cos45o

40 +0 = VA * 0.866 +VB *0.7071

40 = VA * 0.866 +VB *0.7071 ........................eq (i)

By using vertical momentum conservation

momentum before collision = mumentum after collision

0 = mVA sin30o - mVB sin45o

VB sin45o = VA sin30o

VB *0.7071 = VA *0.5 ......................eq(ii)

Put value of VB *0.7071 in eq(i)

40 = VA * 0.866 +   VA *0.5

40 = 1.366  VA

VA  = 40 / 1.366

VA  = 29.28 m/s .....................Ans

Put value VA  = 29.28 in eq (ii)

VB *0.7071 = 29.28 *0.5

VB *0.7071 = 14.64

VB  = 14.64/0.7071

VB = 20.7 m/s ......................Ans

Part(b)

...................................Ans