Chapter 9: Q58P (page 252)

In Fig. 9-62, block 2 (mass 1.0 kg) is at rest on a frictionless surface and touching the end of an un-stretched spring of spring constant . The other end of the spring is fixed to a wall. Block 1 (mass 2.0 kg), traveling at speed $V_{1} = 4.0 m / s$ , collides with block 2, and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed?

Short Answer

Expert verified

Compressed distance of the spring is $x = 0.33 m$

Step by step solution

Step 1: Given Data

Mass of the block 2, $m_{2} = 1 kg$

Initial velocity of block 2, $V_{i 2} = 0 m / s$

Spring constant $k = 200 N / m$

Mass of the block 1, $m_{1} = 2 k g$

Initial velocity of block 1, $v_{i 1} = 4 m / s$

Determining the concept

Usetheprinciple of conservation of momentum and find final velocity ofthetwo blocks. Then, by using conversion of K.E. to spring energy at the point wheretheblocks momentarily stop, find compressed distance x.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formulae are as follow:

$P_{i} = P_{f} P = m v K = \frac{1}{2} m v^{2} K_{s} = \frac{1}{2} k x^{2}$

Determining the compressed distance of the spring (x)

To find compressed distance, find final velocity ofthetwo blocks.Applyingthe principle of conservation of momentum,

Total momentum $\vec{P_{1}}$ before collision = Total momentum after collision role="math" localid="1661491360090" $\vec{P_{f}}$

For the given situation,

Total initial momentum = Initial momentum of block 1+ Initial momentum of block 2.

$\vec{P_{i}} = \vec{P_{i 1}} + \vec{P_{i 2}}$

As initially block e is at rest, $\vec{P_{i 2}} = 0$

$\vec{P_{i}} = \vec{P_{i 1}} = m_{1} v_{i} 1 \cdot \cdot \cdot \cdot \cdot \cdot (1)$

Total final momentum = final momentum of block 1+ final momentum of block 2

As after collision, the blocks stick together, they will have same final velocity.

$\vec{P_{f}} = m_{1} v_{f 1} + m_{2} v_{f 2}$

$v_{f 1} = v_{f 2} = v_{f} \vec{P_{f}} = (m_{1} + m_{2}) v_{f} . . . . . (2)$

Equating equation (1) and (2),

$m_{1} v_{i 1} = (m_{1} + m_{2}) v_{f} v_{f} = \frac{2 \times 4}{2 + 1} v_{f} = 2.66 m / s \approx 2.7 m / s$

Now, using the energy conservation,

$\frac{1}{2} (m_{1} + m_{2}) v^{2} = \frac{1}{2} k x^{2} (1 + 2) 2 . 7^{2} = 200 x^{2}$

Solving this for x,

x=0.33 m.

Hence, the compressed distance of spring will be x=0.33 m.

Therefore, by applying the principle of conservation of momentum and conversion of kinetic energy into spring energy, the compressed distance of the spring can be calculated