Q63P Block 1 of mass m1聽 slides alon... [FREE SOLUTION]

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Chapter 9: Q63P (page 252)

Block 1 of mass $m_{1}$ slides along a frictionless floor and into a one-dimensional elastic collision with stationary block 2 of mass $m_{2} = 3 m_{1}$ . Prior to the collision, the center of mass of the two block system had a speed of 3.00 m/s Afterward, what are the speeds of (a) the center of mass and (b) block 2?

Short Answer

Expert verified

The speeds of the center of mass is, $V_{c o m} = 3.00 m / s$
The speed of the block 2 is, $V_{2 f} = 6.00 m / s$

Step by step solution

Step 1: Given Data

Mass of first block, $m_{1}$

Mass of second block, $m_{2} = 3 m_{1}$

Determining the concept

By using the law of conservation of energy and law of conservation of momentum, solve for velocity of center of mass and speed of block 2. According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formulae are as follow:

For conservation of linear momentum, $m_{1} V_{1 i} + m_{2} V_{2 i} = m_{1} V_{1 f} + m_{1} V_{1 f}$

Conservation of kinetic energy, $\frac{1}{2} m_{1} V_{1 i}^{2} + \frac{1}{2} m_{2} V_{2 i}^{2} = \frac{1}{2} m_{1} V_{1 f}^{2} + \frac{1}{2} m_{2} V_{2 f}^{2}$ .

Speed in center of mass is, $V_{c o m} = \frac{(m_{1} V_{1 i} + m_{2} V_{2 i})}{m_{1} + m_{2}}$

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

Where, m₁, m₂ are masses and V is velocity.

(a) Determining the speed of the center of mass

If external forces are absent, velocity of center of mass does not change. In this problem, no external forces are applied. So, the center of mass velocity before collision and after the collision isthesame.

i.e. $V_{c o m} = 3.00 m / s$

Hence, the speeds of the center of mass is, $V_{c o m} = 3.00 m / s$

(b) Determining the speed of the block

Now, find the velocity $V_{1 i}$ of block 1 before the collision (when the velocity of block 2 is known to be zero),

i.e. $V_{2 i} = 0$

By using equation 9-17,

$V_{c o m} = \frac{(m_{1} V_{1 i} + m_{2} V_{2 i})}{m_{1} + m_{2}} (m_{1} + m_{2}) V_{c o m} = m_{1} V_{1 i} + m_{2} V_{2 i} (m_{1} + m_{2}) V_{c o m} = m_{1} V_{1 i} + 0 (鈭� V_{2 i} = 0) V_{1 i} = \frac{(m_{1} + m_{2}) V_{c o m}}{m_{1}} V_{1 i} = \frac{(m_{1} + 3 m_{1}) V_{c o m}}{m_{1}} V_{1 i} = 4 V_{c o m} V_{1 i} = 12.0 m / s$

Now, to findthe final speed of block 2 ,use equations(9 鈥� 68) inthebook,

$V_{2 f} = \frac{(2 m_{1})}{m_{1} + m_{2}} V_{1 i} V_{2 f} = \frac{(2 m_{1})}{m_{1} + 3 m_{1}} V_{1 i} V_{2 f} = \frac{(2 m_{1})}{4 m_{1}} V_{1 i} V_{2 f} = \frac{V_{1 i}}{2} V_{1 i} = 6.00 m / s$

Hence,the speed of the block 2 is, $V_{2 f} = 6.00 m / s$

Therefore, by using the concept of conservation of mechanical energy, find the velocity after the collision and velocity of center of mass of the system.