A car of mass m0 is moving with a velocity v0 along a straight track. If in a time interval to magnitude of its velocity is increased by 200% and its direction of motion is reversed, then magnitude of change in its momentum in this time interval is equal to n mv0 find n.[Take n positive]
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The angular momentum of mass about center will remain constant as the torque of tension is zero.
mv
∘
r
∘
=mv
′
2
r
∘
v
′
=2v
∘
Final value of K.E.=
2
1
m(2v
∘
)
2
=2mv
∘
2
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