In: Physics
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?
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