A nucleus of mass m originally at rest emits particle of mass m2 with speed 2 m/s. The recoilspeed of the nucleus is(A) 2 m/s (B) 5 m/s(C) 5 m/s (D) 6 m/s

To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the emission of the particle should be equal to the total momentum after the emission.

Let's denote the initial velocity of the nucleus as v1 and the final velocity of the nucleus after the emission as v2. The mass of the nucleus is m and the mass of the emitted particle is m2. The initial velocity of the emitted particle is given as 2 m/s.

Before the emission, the total momentum is given by: Total momentum before = (mass of nucleus) * (initial velocity of nucleus) + (mass of emitted particle) * (initial velocity of emitted particle)

Total momentum before = m * v1 + m2 * 2 m/s

After the emission, the total momentum is given by: Total momentum after = (mass of nucleus) * (final velocity of nucleus) + (mass of emitted particle) * (final velocity of emitted particle)

Total momentum after = m * v2 + m2 * 0 m/s (since the emitted particle has been completely emitted and has no velocity)

According to the principle of conservation of momentum, the total momentum before should be equal to the total momentum after:

m * v1 + m2 * 2 m/s = m * v2

Now, let's solve for v2:

v2 = (m * v1 + m2 * 2 m/s) / m

v2 = v1 + (m2 * 2 m/s) / m

v2 = v1 + 2 m2 / m

Therefore, the recoil speed of the nucleus, v2, is equal to the initial velocity of the nucleus, v1, plus 2 times the initial velocity of the emitted particle divided by the mass of the nucleus.

Now, let's substitute the given values into the equation to find the recoil speed:

v2 = v1 + 2 m2 / m

Since the mass of the emitted particle, m2, is given as m, we can simplify the equation:

v2 = v1 + 2 m / m

v2 = v1 + 2

So, the recoil speed of the nucleus is v2 = v1 + 2 m/s.

Therefore, the correct answer is (A) 2 m/s.