A 6 kg box is travelling on ice at a speed of 9 m/s when a 12 kg packet is gently placed on it. The velocity will now be
Solution:
From Principal of conservation of momentum
m 1 v 1 = (m 1 + m 2 ) v
6 × 9 = ( 6 + 12) × v ⇒ v = 3 m/s
A 6 kg box is travelling on ice at a speed of 9 m/s when a 12 kg packet is gently placed on it. The velocity will now be
From Principal of conservation of momentum
m 1 v 1 = (m 1 + m 2 ) v
6 × 9 = ( 6 + 12) × v ⇒ v = 3 m/s
A ball, moving with a speed v towards north, collides with an identical ball, moving with a speed v towards east. After collision the two balls stick together and move towards north-east. The speed of the combination is
Taking both the balls as one system
mv i + mv j= 2m× v
so, v= v/2 i + v/2 j
so, |v|= v/√2
A bomb of mass M at rest explodes into three pieces, two of which of mass M/4 each, are thrown off in perpendicular directions with speeds of 3 m/s and 4 m/s. The third piece is thrown off with a speed
As the bomb was initially at rest and no external force acts on it total momentum of the bomb should remain constant.
so, (m/4) 3 i +(m/4) 4 j + (m/2) v1 = 0
v1 = 3/2 i + 4/2 j
|V1|= 2.5 m/s