Question

A particle of mass m moving with a velocity v makes an elastic one dimensional collision with a stationary particle of mass m establishing a contact with it for extremely 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 in figure. The magnitude possessed by $F_0$ is :

• A. $\frac{mu}{T}$
• B. $\frac{2mu}{T}$
• C. $\frac{4mu}{3T}$
• D. $\frac{3mu}{4T}$

Answer 1

The correct option is C $\frac{4mu}{3T}$

Initial velocity of the moving body (A) is $v$ and that of body (B) is zero.

The two colliding bodies having same mass exchange their respective velocities after the head-on elastic collision.

Thus velocity of the body (A) after the collision is zero.

$⟹$ Change in momentum of body (A), $\Delta p=mv$

Area under the graph shown, $A=\frac{1}{2}{F}_o\times \frac{T}{4}+F_o\times \frac{T}{2}+\frac{1}{2}{F}_o\times \frac{T}{4}=\frac{3}{4}T{F}_o$

Using $\Delta p=\int Fdt=$ Area under F-t graph

$\therefore$ $mv=\frac{3}{4}T{F}_o$ $⟹$ $F_o=\frac{4mv}{3T}$