Question

A bomb of mass 30kg at rest explodes into two pieces of masses 18kg and 12kg. The velocity of 18kg mass is 6ms-1. The kinetic energy of the other mass is

• A. 243 J
• B. 486 J
• C. 564 J
• D. 388 J

Answer 1

Answer: (B)

486 J

Answer 2

The correct option is(B): 486 J.

The conservation of momentum states that the total momentum before the explosion is equal to the total momentum after the explosion.

Before the explosion, the bomb is at rest, so the total momentum is zero:

Total initial momentum (before explosion) = 0

After the explosion, we have two pieces with masses m1 = 18 kg and m2 = 12 kg. Let v1 be the velocity of the 18 kg mass after the explosion.

Total final momentum (after explosion) = m1 * v1 + m2 * v2

We know that m1 * v1 is the momentum of the 18 kg mass, and since v1 = 6 $\frac{m}{s}$:

Total final momentum = (18 kg) * (6 $\frac{m}{s}$) + (12 kg) * v2

Now, according to the conservation of momentum, the total initial momentum is equal to the total final momentum:

0 = (18 kg) * (6 $\frac{m}{s}$) + (12 kg) * v2

Now, solve for v2:

(12 kg) * v2 = - (18 kg) * (6 $\frac{m}{s}$)

v2 = - (18 kg) * $\frac{(6 m/s) }{ (12 kg)}$

v2 = -9 $\frac{m}{s}$

Now that we have found the velocity of the 12 kg mass (v2 = -9 $\frac{m}{s}$), we can calculate its kinetic energy using the formula for kinetic energy:

Kinetic Energy (KE) = $(\frac{1}{2})$ * mass * velocity2

KE = $(\frac{1}{2})$ * (12 kg) * $(\frac{-9 m}{s^2})$

KE = $(\frac{1}{2})$ * (12 kg) * $\frac{81 m^2}{s^2}$

KE = 486 J.