Question

A proton moving with a velocity of $1.25 \times 10^5 m/s$ collides with a stationary helium atom. The velocity of proton after collision is:
$\text{A.}\quad 0.75 \times 10^5 m/s$
$\text{B.}\quad 7.5 \times 10^5 m/s$
$\text{C.}\quad -7.5 \times 10^5 m/s$
$\text{D.}\quad 0\ m/s$

Answer 1

Collision is an event in which a body exerts a large amount of force on another body for a small interval of time. Collisions can be divided into elastic and inelastic collisions. In elastic collisions, there is no loss in kinetic energy but in non-elastic energy, there is a loss in kinetic energy. Yet momentum is constant in both the collisions.

Formula used:
$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

Complete step-by-step answer:
We are supposed to find the velocity of the proton after the collision happened. We know that the Helium atom consists of two protons and two neutrons. Hence the total mass of the atom is $4\times m$, where ‘m’ is the mass of proton or neutron.
Now, as the momentum remains unchanged, we can apply the principle of conservation of linear momentum.
i.e. $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
Since initially the helium atom was at rest, so $u_2 =0$
Thus $m \times 1.25 \times 10^5 + 0 = m v_1 + 4m \times v_2$
Or $1.25 \times 10^5 = v_1 + 4v_2$
Or $v_2 = \dfrac{1.25 \times 10^5 -v_1}{4}$
Now, by energy conservation, we have;
$\dfrac12 m u^2 = \dfrac12 m v_1^2 + \dfrac12 (4m) v_2^2$
Or $u^2 = v_1^2 + 4 v_2^2$
Now, putting the value of $v_2$, we get;
$(1.25\times 10^5)^2 = v_1^2 + 4\left( \dfrac{1.25 \times 10^5 - v_1}{4} \right)^2$
Or $5v_1^2 - 2.5\times 10^5v_1 - 3 \times (1.25\times 10^5)^2 =0$
Solving for $v_1$ using quadratic formula:
$x = \dfrac{-b \pm \sqrt {b^2 -4ac}}{2a}$
$v_1 = \dfrac{2.5 \times 10^5 \pm \sqrt{(2.5\times 10^5)^2 + 10\times 3\times (1.25\times 10^5)^2}}{2\times 5}$
$v_1 = \dfrac{2.5 \times 10^5 \pm 2.5\times 10^5 \sqrt{1+15}}{10}$ [Taking $(2.5 \times 10^5)$ common inside the root]
$v_1 = \dfrac{2.5 \times 10^5 \pm 10 \times 10^5}{10}$
Hence, $v_1 = 1.25 \times 10^5 m/s \ or \ -0.75 \times 10^5 m/s$
Now, $v_1$ can’t be equal to $1.25 \times 10^5 m/s$ as there must be a change in velocity if the mass of helium is not zero.
Thus, $v_1 = 0.75 \times 10^5 m/s$.

So, the correct answer is “Option A”.

Note: Whenever there is a collision between two objects, the velocity of both bodies must change. It can’t be greater than the original case as described in the question. So, velocity should decrease in case of proton and increase in case of helium. One can directly omit the options B, C and D. As in B and C, the velocity is increased from original and in case of D, it is directly given zero.