Chapter 9: Q67P (page 252)

In Fig.9-66, particle 1 of mass $m_{1} = 0.30 kg$ slides rightward along an x axis on a frictionless floor with a speed of 0.20m/s. When it reaches ,x=0 it undergoes a one-dimensional elastic collision with stationary particle 2 of mass $m_{2} = 0.40 kg$ . When particle 2 then reaches a wall at $x_{w} = 70 cm$ , it bounces from the wall with no loss of speed. At what position on the x axis does particle 2 then collide with particle 1?

Short Answer

Expert verified

The position of collision is -28 cm

Step by step solution

Step 1: Given Data

Mass of particle 1 is, $m_{1} = 0.30 kg$

Initial velocity of mass $m_{1}$ is $V_{1 i} = 2.0 m / s$

Mass ofparticle 2 is, $m_{2} = 0.40 kg$

Distance of wall is, $x_{w} = 70 cm = 0.70 m$

Determining the concept

Fromthelaw of conservation of linear momentum, find the final velocity after the collision. From that velocity, find the final position of collision.According to conservation of linear momentum, momentum that characterizes motion never changes in an isolated collection of objects.

Formulae are as follow:

Conservation of linear momentum is, $m_{1} V_{1 i} + m_{2} V_{2 i} = m_{1} V_{1 f} + m_{2} V_{2 f}$

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

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

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

From equation 9-68, $V_{2 f} = \frac{2 m_{1}}{m_{1} + m_{2}} V_{1 i}$

$V = \frac{x}{t}$

Where, m₁, m₂ are masses, x is displacement, t is time and V is velocity.

Determining the position of collision

Now,

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

Similarly, the total kinetic energy is conserved,

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

Solving equation $V_{1 f}$ is given by equation 9-67,

$V_{1 f} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} V_{1 i} V_{1 f} = \frac{0.30 - 0.40}{0.30 + 0.40} (2.0) V_{1 f} = \frac{- 0.10}{0.70} \times 2.0 V_{1 f} = - 0.29 m / s$

Solving equation $V_{2 f}$ is given by equation 9-68,

$V_{2 f} = \frac{2 m_{1}}{m_{1} + m_{2}} V_{1 i} V_{2 f} = \frac{2 \times 0.30}{0.70} \times 2 V_{2 f} = 1.7 m / s$

When particle 2 is again at position x=0, the total distance covered is $2 x_{w} = 140 cm$ , then calculate the time required for the particle to return again to its position,

$V_{2 f} = \frac{2 x_{w}}{t} t = \frac{2 x_{w}}{V_{2 f}} t = \frac{140 \times 10^{- 2}}{1.7} t = 0.82 s$

At t=0.82 s particleis located at,

$V_{1 f} = \frac{x}{t} x = V_{1 f} t x = - 0.29 \times 0.82 x = 0.23 m$

Particle is gaining at a rate of $\frac{10}{7} m / s$ leftward. This is their relative velocity at that time. Thus, this “gap” of 23 cm between them will be closed after an additional time of $(0.23) / (10 / 7) = 0.16 s$ has passed.