Q42P An object, with mass m and speed... [FREE SOLUTION]

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Chapter 9: Q42P (page 250)

An object, with mass m and speed v relative to an observer, explodes into two pieces, one three times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer鈥檚 reference frame?

Short Answer

Expert verified

The kinetic energy added to the system during the explosion is $K = \frac{1}{6} m v^{2}$ .

Step by step solution

Step 1: Given data:

The mass of the object is m .

The speed of the object is v .

The mass of the first object, $m_{1} = \frac{1}{4} m$

The mass of the second object, $m_{2} = \frac{3}{4} m$

Determining the concept:

By applying conservation of linear momentum, find the velocity and using this value in the equation for increase in the kinetic energy, find thekinetic energy added to the system during the explosion.

Formulae are as follows:

The conservation of linear momentum is,

${\overset{鈫�}{p}}_{f} = {\overset{鈫�}{p}}_{i} m {\overset{鈫�}{v}}_{0} = m_{1} \overset{鈫�}{v_{1}} + m_{2} {\overset{鈫�}{v}}_{2} m v = m_{1} v_{1} + m_{2} v_{2}$

The increase in kinetic energy is,

$K = K_{f} - K_{i} = (\frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}) - (\frac{1}{2} m v_{0}^{2})$

Where, m, m₁, m₂ are masses, , v₁, v₂are velocities and K is change in kinetic energy.

Determining the kinetic energy added to the system during the explosion:

Let, the mass of the original body be $m_{1}$ , its velocity is ${\overset{鈫�}{v}}_{1} = 0$ , the mass of the more massive piece is $m_{2}$ . Note that, $m_{1} + m_{2} = m$ . From this,

$m_{1} = \frac{1}{4} m$ and $m_{2} = \frac{3}{4} m$

The total final momentum of the two pieces is,

$p_{f} = m_{1} v_{1} + m_{2} v_{2}$ 鈥�.. (1)

As less massive piece comes to stop, the velocity of this object is zero. Therefore,

$v_{1} = 0 m / s$

Substitute the known value in the above equation.

$m v = m_{1} (0 m / s) + (\frac{3}{4} m) v_{2} m v = (\frac{3}{4} m) v_{2} v_{2} = \frac{4 v_{0}}{3}$

Applying conservation of linear momentum,

$m {\overset{鈫�}{v}}_{0} = m_{1} \overset{鈫�}{v_{1}} + m_{2} {\overset{鈫�}{v}}_{2} m v \hat{i} = 0 + \frac{3}{4} = m {\overset{鈫�}{v}}_{2} {\overset{鈫�}{v}}_{2} = \frac{4}{3} v \hat{i}$

Therefore, increase in the kinetic energy is given by,

$K = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} - \frac{1}{2} m v_{0}^{2} = 0 + \frac{1}{2} (\frac{3}{4} m) {(\frac{4}{3} v)}^{2} - \frac{1}{2} m v^{2} = \frac{1}{6} m v^{2}$