Chapter 9: Q103P (page 255)

In Fig. 9-80, block 1 of mass $m_{1} = 6.6 k g$ is at rest on a long frictionless table that is up against a wall. Block 2 of mass $m_{2}$ is placed between block 1 and the wall and sent sliding to the left, toward block 1, with constant speed $v_{2 i}$ . Find the value of $m_{2}$ for which both blocks move with the same velocity after block 2 has collided once with block 1 and once with the wall. Assume all collisions are elastic (the collision with the wall does not change the speed of block 2).

Short Answer

Expert verified

The value of mass $m_{2}$ is 2.2 kg .

Step by step solution

Understanding the given information

i) Mass of block 1, $m_{1} i s 6.6 k g$ .

ii) The speed of block 2 is $v_{2 i}$ .

Concept and formula used in the given question

You use the concept of conservation of momentum to find the mass. You use the equations given in the book 9-75 and 9-76 and can solve for the mass of block 2.

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

Calculation for the value of m2 for which both blocks move with the same velocity after block 2 has collided once with block 1 and once with the wall

We can use the below equations to find mass as,

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

Here, $v_{2 i}$ is the initial velocity of block -2.

The velocity of block 2, after bouncing off the wall, will be the same as before but with a negative sign; we can write it as,

$V_{1 f} = - V_{2 f}$

Substitute the values in the above expression, and we get,

$\frac{2 m_{2}}{m_{1} + m_{2}} v_{2 i} = \frac{(m_{2} - m_{1})}{m_{1} + m_{2}} v_{2 i} 2 m_{2} = - m_{2} - m_{1} 3 m_{2} = m_{1} m_{2} = \frac{m_{1}}{3}$