# 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 ...........?

## 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 $\frac{T}{2}$ and decreases linearly from ${F}_{0}$ to zero in further time $\frac{T}{4}$ as shown in figure. The magnitude possesssed by ${F}_{0}$ is ?

Mar 1, 2017

${F}_{o} = \frac{4 m v}{3 T}$

#### Explanation:

Momentum is always conserved so we can say -- where ${u}_{1} , {u}_{2}$ are the post-collision velocities of the first (previously moving) particle and the second (previously stationary) particle, respectively -- that:

$m v = m {u}_{1} + m {u}_{2} \implies v = {u}_{1} + {u}_{2} q \quad \triangle$.

From conservation of (kinetic) energy, we can say:

$\frac{1}{2} m {v}^{2} = \frac{1}{2} m {u}_{1}^{2} + \frac{1}{2} m {u}_{2}^{2} \implies {v}^{2} = {u}_{1}^{2} + {u}_{2}^{2} q \quad \square$

From $\triangle$, we have:

${u}_{2} = v - {u}_{1}$ and so ${u}_{2}^{2} = \left(v - {u}_{1}\right) \left(v - {u}_{1}\right)$

From $\square$, we have:

${u}_{2}^{2} = {v}^{2} - {u}_{1}^{2} = \left(v - {u}_{1}\right) \left(v + {u}_{1}\right)$

$\implies \left(v - {u}_{1}\right) \left(v - {u}_{1}\right) = \left(v - {u}_{1}\right) \left(v + {u}_{1}\right)$

$\implies v = {u}_{1} , {u}_{2} = 0$

or

$\implies v - {u}_{1} = v + {u}_{1} \implies {u}_{1} = 0 , {u}_{2} = v$

From a consideration of the physical setup, we see that the only practical conclusion is that:

${u}_{1} = 0 , {u}_{2} = v$

The first particle cannot just move through the second particle with the second remaining stationary. Rather, all momentum and energy is transferred to the second particle.

We now turn to Netwon's 2nd Law: $F = m a = m \frac{\mathrm{dv}}{\mathrm{dt}}$

Because we have a Force-time graph, we can tweak the second law as follows, by integrating wrt time:

${\int}_{0}^{T} F \setminus \mathrm{dt} = {\int}_{0}^{T} m \frac{\mathrm{dv}}{\mathrm{dt}} \mathrm{dt}$

We do this because we see that ${\int}_{0}^{T} F \setminus \mathrm{dt}$ is the area of that graph, which is simply:

$\frac{1}{2} \cdot \frac{T}{4} \cdot {F}_{o} + \frac{T}{2} \cdot {F}_{o} + \frac{1}{2} \cdot \frac{T}{4} \cdot {F}_{o} = \frac{3}{4} T {F}_{o}$

For the second integration, we change the integration variable (using the chain rule) to get:

${\int}_{0}^{T} m \frac{\mathrm{dv}}{\mathrm{dt}} \mathrm{dt} = m {\int}_{{v}_{1}}^{{v}_{2}} \mathrm{dv}$

$= m \Delta v$

Which here, because ${u}_{2} = v$:

$= m \left(v - 0\right) = m v$

Thus we can say that:

$\frac{3}{4} T {F}_{o} = m v$

Or:

${F}_{o} = \frac{4 m v}{3 T}$