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Grade 11Mechanics

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

Profile image of Aditya
12 Years agoGrade 11
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6 Answers

Profile image of naman sharma
12 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.

Profile image of Lokendra
10 Years ago
The external force present here is gravity. So, change in momentum =F(2t)= (m1+m2)g(2t)
Profile image of Jitesh Singh
9 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
Profile image of tarun
8 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 
Profile image of Aishita batra
8 Years ago
Normally we neglect gravitational force while applying conservation of momentum but we haven`t done so here.why?
Profile image of Harsh gupta
7 Years ago
Fext=(m1+m2)/g=|(m1v1'+m2V2')-(m1v1-m2v2)|/(2t - 0)
so the value of
|(m1v1'+m2V2')-(m1v1-m2v2)|=2(m1+m2)gt