Q59P In Fig. 9-63 , block 1 (mass 2.0... [FREE SOLUTION]

Chapter 9: Q59P (page 252)

In Fig. 9-63 , block 1 (mass 2.0 kg) is moving rightward at 10 m/sand block 2 (mass 5.0 kg) is moving rightward at 3.0 m/s . The surface is frictionless, and a spring with a spring constant of 1120 N/mis fixed to block 2. When the blocks collide, the compression of the spring is maximum at the instant the blocks have the same velocity. Find the maximum compression.

Short Answer

Expert verified

Maximum compression (x) is 25 cm

Step by step solution

Step 1: Given Data

Mass of the block 2, m_{2} = 5 kg Initial velocity of block 2, v_{i 2} = 3 m / s Spring constant k = 1120 N / m Mass of the block 1, m_{1} = 2.0 kg Initial velocity of block 1, v_{i 1} = 10 m / s

Determining the concept

Usetheprinciple of conservation of momentum and find final velocity of two blocks. Then, by using conversion of K.E. to spring energy at maximum compression of the spring, 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:

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

where, m is mass, v is velocity, P is linear momentum, x is displacement, k is spring constant and K is kinetic energy.

Determining the maximum compression (x)

To find compressed distance, findthefinal velocity of two blocks.Applying principle of conservation of momentum,

Total momentum $\overset{鈬赌}{P_{i}}$ before collision = Total momentum after collision $\overset{鈬赌}{P_{f}}$

For the given situation

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

$\overset{鈬赌}{P_{i}} = \overset{鈬赌}{P_{i 1}} + \overset{鈬赌}{P_{i 2}} \overset{鈬赌}{P_{i}} = m_{1} v_{i 1} + m_{2} v_{i 2} . . . . . . . (1)$

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

It is given that at the maximum compression blocks will have same final velocity,

$\overset{鈬赌}{P_{i}} = m_{1} v_{f 1} + m_{2} v_{f 2} v_{f 1} = v_{f 2} = v_{f} \overset{鈬赌}{P_{i}} = (m_{1} + m_{2}) v_{f} . . . . . . . (2)$

Equating equation (1) and (2),

$m_{1} v_{i 1} + m_{2} v_{i 2} = (m_{1} + m_{2}) v_{f} v_{f} = \frac{m_{1} v_{i 1} + m_{2} v_{i 2}}{(m_{1} + m_{2})} v_{f} = \frac{2 脳 10 + 5 脳 3}{2 + 5} v_{f} = 5 m / s$

Now, to find out change in K.E. energy of the system,

$K_{i} = \frac{1}{2} m_{1} v_{i 1}^{2} + \frac{1}{2} m_{2} v_{i 2}^{2} K_{i} = \frac{1}{2} (m_{1} + m_{2}) v_{f}^{2} 鈭� K = K_{f} - K_{i} 鈭� K = \frac{1}{2} (m_{1} + m_{2}) v_{f}^{2} - \frac{1}{2} m_{1} v_{i 1}^{2} + \frac{1}{2} m_{2} v_{i 2}^{2} . . . . . . . (3)$

At maximum compression x in the spring, change in K.E. will get stored in to the spring,

$鈭� K = \frac{1}{2} k x^{2} . . . . . . . . (4)$

Using equation (3) and (4),

$\frac{1}{2} (m_{1} + m_{2}) v_{f}^{2} - \frac{1}{2} m_{1} v_{i 1}^{- 2} \frac{1}{2} m_{2} v_{i 2}^{- 2} = \frac{1}{2} k x^{2}$

Cancelling 陆 from both sides,

$k x^{2} = (m_{1} + m_{2}) v_{f}^{2} - m_{1} v_{i 1}^{2} - m_{2} v_{i 2}^{2} k x^{2} = (2 + 5) 5^{2} - 2 脳 10^{2} - 5 脳 3^{2} x^{2} = \frac{- 70}{1120} = - . 0.0625$

Taking square root and considering positive value for distance,

$x = \sqrt{0.0625} x = 0.25 m x = 25 cm$

Hence, the compressed distance of spring will be 25 cm.

Therefore, by applying the principle of conservation of momentum and conversion of K.E. to spring energy we have calculated the compressed distance of the spring.