Correct Answer - Option 4 : 2 : 1
CONCEPT:
Linear momentum:
- It is defined as the product of the mass of an object, m, and its velocity, v. It, is a vector quantity.
⇒ P = mv
Where m = mass, and v = velocity
Conservation of momentum:
- If the force applied to the system of particles is zero, then the total momentum of the system of particles will remain conserved.
⇒ P = P1 + P2 + P3 +...+Pn = constant
Kinetic energy
- The energy possessed by a body due to the virtue of its motion is called kinetic energy.
\(⇒ KE=\frac{1}{2}mv^{2}\)
CALCULATION:
Given m = initial mass of the bomb, \(\frac{m_1}{m_2}=\frac{1}{2}\), and (velocity of m2) v2 = v
Initially, the bomb is at rest so the initial velocity of the bomb is,
⇒ v = 0 m/sec
- The initial momentum of the system is given as,
⇒ Pi = mv
⇒ Pi = m × 0
⇒ Pi = 0 kg-m/sec -----(1)
⇒ \(\frac{m_1}{m_2}=\frac{1}{2}\)
⇒ m1 = m' and ∴ m2 = 2m'
Let the velocity of m1 be v1.
- So the final momentum of the system is given as,
⇒ Pf = m1v1 + m2v2
⇒ Pf = m'v1 + 2m'v -----(2)
Since here no external force is acting on the bomb, so according to the law of conservation of momentum,
⇒ Pf = Pi
⇒ m'v1 + 2m'v = 0
⇒ v1 = -2v -----(3)
The kinetic energy of the lighter part is given as,
\(⇒ KE_1=\frac{1}{2}m_1v_1^{2}\)
\(⇒ KE_1=\frac{1}{2}m'(2v)^{2}\)
⇒ KE1 = 2m'v2 -----(4)
The kinetic energy of the heavier part is given as,
\(⇒ KE_2=\frac{1}{2}m_2v_2^{2}\)
\(⇒ KE_2=\frac{1}{2}(2m')v^{2}\)
⇒ KE2 = m'v2 -----(5)
By equation 4 and equation 5,
\(⇒ \frac{KE_1}{KE_2}=\frac{2m'v^2}{m'v^2}\)
\(⇒ \frac{KE_1}{KE_2}=\frac{2}{1}\)
- Hence, option 4 is correct.