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two bodies of mass 1 kg and 2 kg move towards each other in mutually perpendicular direction with velocities 3m/s and 2 m/s respectively. If the bodies stick together after collision the heat liberated will be ? I.need the process of doing

two bodies of mass 1 kg and 2 kg move towards each other in mutually perpendicular direction with velocities 3m/s and 2 m/s respectively. If the bodies stick together after collision the heat liberated will be ? I.need the process of doing

Grade:12

5 Answers

Akash Kumar Dutta
98 Points
8 years ago

Dear Shuvam,
total momentum before collision=7 units.
after collision M=3 so v=7/3 units,
final energy=1/2Mv^2=49/6 J
initial energy= 1/2m1(v1)^2 + 1/2 m2.(v2)^2
                    = 9/2+4=17/2 J
Loss in energy= initial - final=17/2 - 49/6=2/6=1/3= .33 Joules(ANS)

Regards.

tejas naresh patel
40 Points
8 years ago
60
Shuvam Shukla
37 Points
8 years ago

Ans is 13/3...

Ansh Mahapatra
40 Points
2 years ago
By law of conservation of momentum,
1*3+2*2=3*v
We get,v=7/3 m/s
Initial energy=17/2 J
Final energy=49/6 J
Energy lost as heat=1/3 J=0.33 J
Aditya
15 Points
2 years ago
Answer is 13/3 because if we conserve momentum and subtract it we get the answer.

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