A neutron having mass of 1.67 x 10⁻²⁷ kg and moving at 10⁸ m / s collides with a deutron at rest and sticks to it. If the mass of the deutron is 3.34 x 10⁻²⁷kg; the speed of the combination is

The correct option is 3.33 x 10⁷ m / s.

Step-by-Step Explanation:

We are given the following values:
Mass of the neutron, $m_1=1.67\times {10}^{-27}\text{kg}$
Initial velocity of the neutron, $u_1={10}^8\text{m/s}$
Mass of the deuteron, $m_2=3.34\times {10}^{-27}\text{kg}$
Initial velocity of the deuteron (since it is at rest), $u_2=0\text{m/s}$

When the neutron collides with the deuteron and sticks to it, this is a completely inelastic collision. Let the final common speed of the combined mass (neutron + deuteron) be $v$.

According to the law of conservation of linear momentum:
Total initial momentum = Total final momentum
$m_1{u}_1+m_2{u}_2=(m_1+m_2)v$

Substituting the given values into the equation:
$(1.67\times {10}^{-27}\text{kg})\times \left({10}^8\text{m/s}\right)+0=(1.67\times {10}^{-27}\text{kg}+3.34\times {10}^{-27}\text{kg})\times v$

Simplifying the right side:
$1.67\times {10}^{-19}=(5.01\times {10}^{-27})\times v$

Solving for the common speed $v$:
$v=\frac{1.67\times {10}^{-19}}{5.01\times {10}^{-27}}$
$v=\frac{1.67}{5.01}\times {10}^8\text{m/s}$
$v=\frac{1}{3}\times {10}^8\text{m/s}$
$v\approx 0.333\times {10}^8\text{m/s}=3.33\times {10}^7\text{m/s}$

Thus, the final speed of the combination is 3.33 x 10⁷ m / s.