Chapter 9: Q62P (page 252)

Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the spheres, whose mass is 300 g , remains at rest. (a) What is the mass of the other sphere? (b) What is the speed of the two-sphere center of mass if the initial speed of each sphere is 2.00 m/s?

Short Answer

Expert verified

The mass of the other sphere $m_{2} = 100 g$
The speed of the two sphere center of mass is, $V_{c o m} = 1.00 m / s$

Step by step solution

Step 1: Given Data

Mass of one sphere, $m_{1} = 300 g$

After collision, speed of each sphere is, $V_{1 f} = V_{2 f} = 2 m / s$

Determining the concept

According to the law of conservation of linear momentum, if the system is closed and isolated, the total linear momentum of the system must be conserved. If the bodies stick together, the collision is a completely inelastic collision. In an inelastic collision, kinetic energy is not conserved.

Formulae are as follow:

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

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

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

(a) Determining the mass of the other sphere

Letbe the mass of the sphere 1, $V_{1 i}$ and $V_{1 f}$ be its velocities before and after the collision respectively. Let $m_{2}$ be the mass of the second sphere, $V_{2 i}$ and $V_{2 f}$ be its velocity before and after the collision momentum.

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

Let’s assume that the direction of motion of the first sphere is positive and it comes to rest after the collision. So, sphere 2 would be traveling along negative direction. Hence, write $V_{1 i}$ ,as V ,and $V_{2 i}$ with -V, and $V_{1 f}$ equal to zero to obtain $0 = m_{1} - 3 m_{2}$ .

$\begin{array}{l} 0 = m_{1} - 3 m_{2} \\ m_{2} = \frac{m_{1}}{3} \\ m_{2} = \frac{300}{3} \\ m_{2} = 100 g \end{array}$

Hence, the mass of the other sphere is, $m_{2} = 100 g$ .

(b) Determining the speed of the two spheres’ center of mass

Use the velocities before the collision to calculate the velocity of the center of mass.

Speed in center of mass is,

$V_{c o m} = \frac{(m_{1} V_{1 i} + m_{2} V_{2 i})}{m_{1} + m_{1}} V_{c o m} = \frac{(300 \times 2 + 100 \times (- 2))}{300 + 100} V_{c o m} = \frac{600 - 200}{400} V_{c o m} = 1.00 m / s$

Hence,the speed of the two sphere center of mass is, $V_{c o m} = 1.00 m / s$

Therefore, by using the concept of conservation of liner momentum, the mass of the second sphere before the collision and velocity of center of mass of the system can be found.