Asked by triumphant
A body of mass 4.2kg moving with a velocity of 10m/s due east,hit a stationary body of mass 2.8kg.if they stick together after collision and move with a velocity v due east calculate the value of v
Answers
Answered by
bobpursley
stick together: only law of conservation of momentum applies
4.2*10E+2.8*OE= (4.2+2.8)V
V= 42/(7.0) East
4.2*10E+2.8*OE= (4.2+2.8)V
V= 42/(7.0) East
Answered by
henry2,
M1 = 4.2kg, V1 = 10 m/s.
M2 = 2.8kg, V2 = 0.
Momentum before = Momentum after
M1*V1 + M2*V2 = M1V + M2*V.
4.2*10 + 2.8*0 = 4.2*V + 2.8*V,
V =
M2 = 2.8kg, V2 = 0.
Momentum before = Momentum after
M1*V1 + M2*V2 = M1V + M2*V.
4.2*10 + 2.8*0 = 4.2*V + 2.8*V,
V =
Answered by
Idera
M1U=(m1+m2)v
=4.2*10=(4.2+2.8)v
v=4.2×10/7
v=6m/s
=4.2*10=(4.2+2.8)v
v=4.2×10/7
v=6m/s
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