Hello students, in this article we are going to deal with the important article i.e. conservation of momentum. Momentum of two interacting bodies always remains conserved.
The law states that ‘if no external force acts on interacting bodies, their momentum remains conserved.’ Or the ‘total momentum of two colliding bodies before and after collision is constant.’
To understand this consider the following example,
Suppose two persons of different masses m1 and m2 are moving with velocities of u1 and u2 (u1>u2) while jogging. Imagine that the path is narrow, they get collide on each other, their motion gets affected due to this and now they moves with velocities v1 and v2.
In this case, before collision, the total momentum of person A and B will be
P1= m1u1+ m2u2 …………………..(A)
After collision the momentum will be
P2= m1v1+m2v2 …………………..(B)
During to collision, person A exerts force F1 on person B; in return according to Newton’s 3rd law person B will also exert force F2 on person A in time ‘t’. These forces are equal in magnitude but opposite in direction, hence
Following are the examples where the law of conservation of momentum is obeyed.
- Launching of satellite
- When a person in boat jumps into water, boat gets slightly jerked
- Fire crackers
Ex:1) A boy running with speed of 5 m/s has mass 40 kg, strikes on a stone of mass 2 kg on road which was at rest earlier. After collision boy fumbled and moved with speed of 4.8 m/s, what should be velocity stone which moved it?
Ans: Here m1 = 40 kg , m2 = 2 kg, u1 = 5 m/s, u2 = 0 m/s, v1 = 4.8 m/s, v2 =?
By law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
40 × 5 + m2 × 0 = 40 × 4.8 +2v2
200 = 192 + 2v2
2v2 = 200 – 192 = 8
v2 = 4 m/s
Hence stone will moved with velocity of 4 m/s.
Ex:2) Krishna throws an arrow of mass 2 kg with velocity of 20 m/s pick the fruit of mass 100 g from the tree. After collision fruit and arrow both falls on ground. Find velocity of arrow and fruit before falling on ground.
Ans: Here m1 = 2 kg , m2 = 100 g = 0.1 kg, u1 = 20 m/s, u2 = 0 m/s,
v1 = v2 (since the fruit is stucked at tip of arrow) = ?
By law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
2 × 20 + m2 × 0 = m1v1 + m2v1
40 = (m1 +m2)v1 = (2+0.1)v1
40=(2.1)v1
v1 = 40/2.1≈ 19 m/s
Hence arrow and fruit will moved with velocity of 19 m/s before falling on ground.
Very nice questions