Question

By: Munesh Barua
1. TWO objects, each of mass 1.5 kg are moving in the same straight line but in opposite direction.
The velocity of each object is 2.5 m/s before the collision during which they stick together. What
will be the velocity of the combined object after collision?
2. A bullet of mass 10 gm travelling horizontally with a velocity of 150 m/s strikes a stationary
wooden block and comes to rest in 0.03 sec. calculate the distance of penetration of the bullet
into the block?
3. A stone of mass 1 kg thrown with a velocity of 20m/s across the frozen surface of a lake and
comes to rest travelling a distance of 50m. what is the force of friction between stone and the
ice?
4. A body of mass 300 gm kept at rest breaks into two parts due to internal force. One part of 200
gm is found to move at a speed of 12 m/s towards the east. What will be the velocity of the
other part?
5. Two blocks of mass m, and m2 (m = 1 kg and m:=2 kg) are placed in contact on a frictionless
horizontal surface. A force of 10 N is acting on m What is the acceleration produced in mand
m2?
6. A boy weighing 30 gm is riding bicycle welghing 50 kg. if the bicycle is moving with the speed of
9 km/h towards the west, find the linear momentum of the bicycle-boy system.
I l eden bil from the top of a tower that is 100 m high and at the same time​

Answer 1

Answer:

2.5m/s

Explanation:

let object of masses m1 and m2

their initial velocity be u1 and u2

and,their final velocity be v1 and v2

therefore,total momentum before collusion:

m1×u1 + m2×u2

=1.5×2.5 +1.5×2.5

=3.75 +3.75

=7.5

total momentum after collusion:

m1×v1 + m2×v2

=1.5×v +1.5×v

=1.5v+1.5v

=3v

according to law of conservation of momentum:

total momentum before collusion=total momentum after collusion.

7.5=3v

=

$v = \frac{7.5}{3} = \frac{75}{3 \times 10} = 2.5$

therefore,final velocity is 2.5 m/s