A ball of 0.4 kg mass and a speed of 3 m/s has a head-on, completely elastic collision with a 0.6-kg mass initially at rest. Find the speeds of both bodies after the collision.
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By conservation of momentum,
m1v1 + m2v2 = m1u1 + m2u2
m1 = 0.4
m2= 0.6
v1= 3
v2 =0
0.4*3 + 0 = 0.4u1 + 0.6 u2
u1= (1.2-0.6u2)/0.4
By conservation of kinetic energy
m1v1^2 = m1u1^2 + m2u2^2
0.4*9 = 0.4u1^2 + 0.6u2^2
put the value of u1 here, you will get the value of u2...
then you can find the answer for u1
m1v1 + m2v2 = m1u1 + m2u2
m1 = 0.4
m2= 0.6
v1= 3
v2 =0
0.4*3 + 0 = 0.4u1 + 0.6 u2
u1= (1.2-0.6u2)/0.4
By conservation of kinetic energy
m1v1^2 = m1u1^2 + m2u2^2
0.4*9 = 0.4u1^2 + 0.6u2^2
put the value of u1 here, you will get the value of u2...
then you can find the answer for u1
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