Chapter 9: Q65P (page 252)

A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body center of mass if the initial speed of the 2.0 kgbody was 4.0 m/s?

Short Answer

Expert verified

Mass of another body is, $m_{2} = 1.2 kg$
The speed of the two-body center of mass is, $V_{c o m} = 2.5 m / s$ .

Step by step solution

Step 1: Given Data

Mass of first body 1 is, $m_{1} = 2.0 kg$

Final speed of first body is, $V_{1 f} = \frac{V_{1 i}}{4}$

Initial velocity of body 1 is, $V_{1 i} = 4.0 m / s$

Initial velocity of body 2 is, $V_{1 i} = 0 m / s$

Determining the concept

In an elastic collision, the total energy and linear momentum must be conserved. From the law of conservation of linear momentum, find the mass of another object.According to conservation of linear momentum, momentum that characterizes motion never changes in an isolated collection of objects.

Formulae are as follow:

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

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

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

For elastic collision, $V_{1 f} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} V_{1 i}$

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

(a) Determining the mass of the body

Let,andare masses of two bodies respectively, where bodyis at rest. Using the law of conservation of momentum,

$m_{1} V_{1 i} = m_{1} V_{1 f} + m_{2} V_{2 f} V$

Similarly, the total kinetic energy is conserved,

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

Solving equationis given by equation 9-67,

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

Solving for,

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

The given condition is $V = \frac{V_{1 i}}{4}$ , find the second mass to be,

$m_{2} = (\frac{V_{1 i} - \frac{V_{1 i}}{4}}{V_{1 i} + \frac{V_{1 i}}{4}}) m_{1} m_{2} = \frac{3}{5} m_{1} m_{2} = \frac{3}{5} \times 2.0 m_{2} = 1.2 kg$

Hence,mass of another body is, $m_{2} = 1.2 kg$

(b) Determining the speed of the two-body center of mass if the initial speed of the 2.0 kg body was

The speed of the center of mass is,

$V_{c o m} = \frac{(m_{1} V_{1 i} + m_{2} V_{2 i})}{m_{1} + m_{2}} V_{c o m} = \frac{(2.0 \times 4 + 1.2 \times 0)}{2.0 + 1.2} V_{c o m} = \frac{8.0}{3.2} V_{c o m} = 2.5 m / s$