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A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is 6 m s$^{-1}$. The kinetic energy of the other mass is :
A
524 J
B
256 J
C
486 J
D
324 J
Detailed Solution
The bomb is at rest, so its initial momentum is zero. By conservation of momentum the two pieces have equal and opposite momenta.
Momentum of the 18 kg piece $= 18 \times 6 = 108$ kg m/s
So the 12 kg piece also has momentum $p = 108$ kg m/s (in the opposite direction).
Kinetic energy: $KE = \dfrac{p^2}{2m}$
$KE = \dfrac{(18 \times 6)^2}{2 \times 12} = \dfrac{11664}{24}$
$KE = 486$ J
(The velocity of the 12 kg piece is $\dfrac{108}{12} = 9$ m/s, and $\dfrac{1}{2} \times 12 \times 9^2 = 486$ J.)
Momentum of the 18 kg piece $= 18 \times 6 = 108$ kg m/s
So the 12 kg piece also has momentum $p = 108$ kg m/s (in the opposite direction).
Kinetic energy: $KE = \dfrac{p^2}{2m}$
$KE = \dfrac{(18 \times 6)^2}{2 \times 12} = \dfrac{11664}{24}$
$KE = 486$ J
(The velocity of the 12 kg piece is $\dfrac{108}{12} = 9$ m/s, and $\dfrac{1}{2} \times 12 \times 9^2 = 486$ J.)