A particle of mass `m` moving with a velocity `u` makes an elastic one-dimensional collision with a stationary particle of mass `m` establishing a contact with it for extermely small time. `T`. Their force of contact increases from zero to `F_0` linearly in time `T//4`, remains constant for a further time `T//2` and decreases linearly from `F_0` to zero in further time `T//4` as shown. The magnitude possessed by `F_0` is..A. `("mu")/T`B. `(2"mu")/T`C. `(4"mu")/(3T)`D. `(3"mu")/(4T)`

A particle of mass `m` moving with a velocity `u` makes an elastic one

1.	A particle of mass `m` moving with a velocity `u` makes an elastic one-dimensional collision with a stationary particle of mass `m` establishing a contact with it for extermely small time. `T`. Their force of contact increases from zero to `F_0` linearly in time `T//4`, remains constant for a further time `T//2` and decreases linearly from `F_0` to zero in further time `T//4` as shown. The magnitude possessed by `F_0` is..A. `("mu")/T`B. `(2"mu")/T`C. `(4"mu")/(3T)`D. `(3"mu")/(4T)`
Answer» Correct Answer - C Impulse `=` area of tapezium, `=1/2(T+T/2)F_(0)=(3TF_(0))/4` According to impulse momentum theorem. Impulse change in momentum `implies (3TF_(0))/4="mu"impliesF_(0)=(4"mu")/(3T)`

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