KCET · Physics · Center of Mass Momentum and Collision
A shell of mass \(20 \mathrm{~kg}\) at rest explodes into two fragments whose masses are in the ratio \(2: 3\). The smaller fragment moves with a velocity of \(6 \mathrm{~ms}^{-1}\). The kinetic energy of the larger fragment is
- A \(96 \mathrm{~J}\)
- B \(216 \mathrm{~J}\)
- C \(144 \mathrm{~J}\)
- D \(360 \mathrm{~J}\)
Answer & Solution
Correct Answer
(A) \(96 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
Total mass of the shell \(=20 \mathrm{~kg}\)
Ratio of the masses of the fragments \(=2: 3\)
\(\therefore\) Masses of the fragments are \(8 \mathrm{~kg}\) and \(12 \mathrm{~kg}\)
Now, according to the conservation of momentum
\(m_{1} v_{1}=m_{2} v_{2}\) \(\therefore \quad 8 \times 6=12 \times v\) \(v(\) velocity of the larger fragment) \(=4 \mathrm{~m} / \mathrm{s}\) Kinetic energy \(=\frac{1}{2} m v^{2}=\frac{1}{2} \times 12 \times(4)^{2}=96 \mathrm{~J}\)
Kinetic energy \(=\frac{1}{2} m v^{2}=\frac{1}{2} \times 12 \times(4)^{2}=96 \mathrm{~J}\)
Ratio of the masses of the fragments \(=2: 3\)
\(\therefore\) Masses of the fragments are \(8 \mathrm{~kg}\) and \(12 \mathrm{~kg}\)
Now, according to the conservation of momentum
\(m_{1} v_{1}=m_{2} v_{2}\) \(\therefore \quad 8 \times 6=12 \times v\) \(v(\) velocity of the larger fragment) \(=4 \mathrm{~m} / \mathrm{s}\) Kinetic energy \(=\frac{1}{2} m v^{2}=\frac{1}{2} \times 12 \times(4)^{2}=96 \mathrm{~J}\)
Kinetic energy \(=\frac{1}{2} m v^{2}=\frac{1}{2} \times 12 \times(4)^{2}=96 \mathrm{~J}\)
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