A block of mass $m_{1} = 2 k g$ slides on a frictionless table with speed of $10 m / s$ . In front of it, another block of mass $m_{2} = 5 k g$ is moving with speed $3 m / s$ in the same direction. A massless spring of spring constant $k = 1120 N / m$ is attached on the backside of m_{2} as shown. The maximum compression of the spring (in $c m$ ) when the blocks collide is

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Solution

Find final speed of the block of mass $m_{1}$ .
Given, mass of first block, $m_{1} = 2 k g$
Mass of another block, $m_{2} = 5 k g$
Initial speed of first block, $v_{1} = 10 m / s$
Initial speed of the second block, $v_{2} = 3 m / s$
Let at maximum compression both the blocks move with velocity $V$
By conservation of momentum

$V = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}} = 5 m / s$

Find maximum compression in spring.
The change in $K . E . = k_{f} - k_{i}$

$Δ K . E . = \frac{1}{2} (m_{1} + m_{2}) V^{2} - \frac{1}{2} m_{1} v_{1}^{2} - \frac{1}{2} m_{2} v_{2}^{2}$

$Δ K . E . = - 35 J$
This is stored as spring $P E$ .
Therefore $\frac{1}{2} k x^{2} = Δ K \Rightarrow x = \sqrt{\frac{2 Δ K}{k}}$

on solving $x = 0.25 m = 25 c m$
Final answer: 25

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A block of mass $m_{1} = 2 k g$ slides along a frictionless table with a speed of 10m/s. Directly in front of it, and moving in same direction is a block of mass $m_{2} = 5 k g$ moving at a speed of 3 m/s. A massless spring with a constant of k = 1120 N/m is attached to the back side of $m_{2}$ as shown. When the block collides, what is the maximum compression in the spring?