You visited us 1 times! Enjoying our articles? Unlock Full Access!

Velocity of Separation and Approach

Three blocks ...

Question

Three blocks are placed on a smooth horizontal surface and lie on the same horizontal straight line. Block $1$ and block $3$ have mass $m$ each and block $2$ has mass $M (M > m)$ . Block $2$ and block $3$ are initially stationary, while block $1$ is initially moving towards block $2$ with speed $v$ as shown. Assume that all collisions are head-on and perfectly elastic. What value of $M / m$ ensures that block $1$ and block $3$ have the same final speed?

5 + \sqrt{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 - \sqrt{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 + \sqrt{5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3 + \sqrt{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $2 + \sqrt{5}$
Initially collision occurs between block-1 and block-2
Let $v_{1}$ and $v_{2}$ are velocities after collision for block -1 and block-2 respectively.
Consider $a = M / m$ .
Due to momentum conservation,
$m v = m v_{1} + M v_{2}$ or $v = v_{1} + a v_{2}$ ...(1)

Due to energy conservation,
$\frac{1}{2} m v^{2} = \frac{1}{2} m v_{1}^{2} + \frac{1}{2} m v_{2}^{2} + \frac{1}{2} M v_{2}^{2}$ or $v^{2} = v_{1} + a v_{2}^{2}$ ...(2)

By solving (1) and (2), we get
$V_{1} = \frac{1 - α}{1 + α} v$ ...(3)

$v_{2} = \frac{2}{1 + α} v$ ...(4)

Now block-2 collides with block-3.
Let $v_{3}$ and $v_{4}$ are velocities after collisions of block-2 and block-3 respectively.

By Conservation of Momentum,
$M v_{2} = M v_{3} + m v_{4}$ or $a v_{2} = a v_{3} + v_{4}$ or $a_{4} a (v_{2} - v_{3})$ ...(5)

By Conservation of energy,
$\frac{1}{2} M v_{2}^{2} = \frac{1}{2} M v_{3}^{2} + \frac{1}{2} m v_{4}^{2}$ or $a v_{2}^{2} = a v_{3}^{2} + v_{4}^{2}$ ....(6)

using eqns. (5) and (6), we get quadratic eqn.
$(1 + α) \times v_{3}^{2} - (2 \times α \times v_{2}) \times v_{3} + (α - 1) \times v_{2}^{2}$ ...(7)

Two solutions of the above quadratic equation are, $v_{2} = v_{2}$ and $v_{2} = \frac{α - 1}{α + 1} v_{2}$
the solution $v_{2} = v_{2}$ is unacceptable becuase it gives $v_{4} = 0$ . Hence if we consider the other solution, we get $v_{4} = \frac{4 a}{(1 + a)_{2}}$
Now if we equate the magnitude of $v_{4}$ and $v_{1}$ , we get $\frac{4 α}{(1 + α) 2} = \frac{α - 1}{1 + α}$ ...(8)
[it is to be noted if $a > 1$ , sign of $v_{1}$ is negative, hence we copared the magnitude of $v_{1}$ and $v_{4}$ in eqn.(8)]
solving eqn.(8) for $α$ , we get an aceeptable solution. $a = 2 + \sqrt{5}$ or $(M / m) = 2 + \sqrt{5}$

Suggest Corrections

Similar questions

m_{1}

m_{2}

m_{1}

m_{2}

M_{1}, M_{2}, M_{3}

M_{4}

M_{1} >> M_{2} >> M_{3} >> M_{4} .

M_{1}

v_{0}

M_{2} .

M_{4}

M

5 m s^{- 1}

M

2 m

0.25

g = 10 m s^{- 2}

1

3

2

0.5 {m/s}^{2}

m

2 m

2 m

g = 9.81 {m/s}^{2}

Join BYJU'S Learning Program