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Two particles of masses m1 and m2 in projectile motion have velocities v1 and v2 resp. at time t=0. They collide at t=t1. There velocities become v''1 and v''2 at time 2t1 while still moving in air. The value of (m1v''1+m2v''2)-(m1v1+m2v2) is

Two particles of masses m1 and m2 in projectile motion have velocities v1 and v2 resp. at time t=0. They collide at t=t1. There velocities become v''1 and v''2 at time 2t1 while still moving in air. The value of (m1v''1+m2v''2)-(m1v1+m2v2) is

Grade:11

6 Answers

naman sharma
37 Points
7 years ago

First write V1 ad v2 etc i vector form .
Now since during collision only iternal forces act so momentum will be conserved SO THE ANSWER TO YOUR QUESTIO IS 0.

Lokendra
11 Points
5 years ago
The external force present here is gravity. So, change in momentum =F(2t)= (m1+m2)g(2t)
Jitesh Singh
21 Points
4 years ago
talking in vectors we have,v1`=v1+gt0andv2`=v2 +gt0.there fore, value of(m1v``1+m2v``2)-(m1v1+m2v2) is equal to {m1(v1`-v1) + m2(v2`-v2)}.and so we have 2(m1+m2)gt0.as our final answer
tarun
21 Points
3 years ago
in this question we will use the formula 
change in momentum = force * time
since gravity is acting so force = (M1 + M2)g
hence we can write
(M1 + M2)g*2t 
Aishita batra
11 Points
3 years ago
Normally we neglect gravitational force while applying conservation of momentum but we haven`t done so here.why?
Harsh gupta
13 Points
2 years ago
Fext=(m1+m2)/g=|(m1v1'+m2V2')-(m1v1-m2v2)|/(2t - 0)
so the value of
|(m1v1'+m2V2')-(m1v1-m2v2)|=2(m1+m2)gt

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