Name: Power Transmitted by a Wave on a String
Uploaded: 2019-07-19
Duration: 7 min 4 s
Description: We know that a wave transfers energy from one point to another, but to find the power transmitted by a wave on a string, watch this video.

Question

Prove that the maximum velocity imparted by an

α

-particle to a proton during their collision is

1.6

of the initial velocity of the

α

-particle.

Accepted Answer

Let the proton be initially at rest relative to a stationary reference frame, and let the $α$ -particle have a velocity $v_{0}$ . The process of their elastic collision is described by the momentum conservation law

$4 m v_{0} = m v_{1} + 4 m v_{2}$

and by the energy conservation law

$\frac{4 m v_{0}^{2}}{2} = \frac{m v_{1}^{2}}{2} + \frac{4 m v_{2}^{2}}{2},$

where $v_{1}$ and $v_{2}$ are the velocities of the proton and the $α$ -particle in the stationary reference frame after the collision, and $m$ and $4 m$ are the masses of the proton and the $α$ -particle respectively.

Let us consider the collision of these particles in the centre-of-mass system, i.e. in an inertial reference frame moving relative to the stationary reference frame at a velocity

$v^{'} = \frac{4 m v_{0}}{m - 4 m} = \frac{4}{5} v_{0}$

(the numerator of the first fraction contains the total momentum of the system, and the denominator contains its total mass). Figure shows the

velocity $v_{0}$ and the velocities of the $α$ -particle (vector $\vec{O B}$ ) and the proton (vector $\vec{O A}$ ) in the centre-of-mass system before the collision: $O B = (\frac{1}{5}) v_{0}$ and $O A = (\frac{4}{5}) v_{0}$ . According to the momentum conservation law, after the collision, the velocity vectors $\vec{O B}$ and $\vec{O A}$ of the $α$ -particle and the proton must lie on the same straight line, and the relation $O B^{'} : O A^{'} = 1 : 4$ (see Fig.) must be satisfied. According to the energy conservation law, $O B^{'} = O B$ and $O A^{'} \Rightarrow O A$ (prove this).

In the stationary reference frame, the velocities of the $α$ -particle and the proton are represented in the figure by vectors ${\vec{O C}}_{2} = \vec{O B^{'}} + v^{'}$ and $\vec{O C_{1}} = \vec{O A^{'}} + v^{'}$

In order to solve the problem, we must determine the maximum possible length of vector $\vec{O C}$ , i.e. in the isosceles triangle $O A^{'} C_{1}$ , we must determine the maximum possible length of the base for constant values of the lateral sides. Obviously, the maximum magnitude of ${\vec{O C}}_{1}$ is equal to $2 \times (\frac{4}{5}) v_{0} = 1.6 v_{0}$ . This situation corresponds to a central collision.

Prove that the maximum velocity imparted by an $α$ -particle to a proton during their collision is $1.6$ of the initial velocity of the $α$ -particle.

Important Questions on Mechanics

Learn and Prepare for any exam you want !

Prove that the maximum velocity imparted by an α-particle to a proton during their collision is 1.6 of the initial velocity of the α-particle.

Important Questions on Mechanics

Learn and Prepare for any exam you want !

Prove that the maximum velocity imparted by an $α$ -particle to a proton during their collision is $1.6$ of the initial velocity of the $α$ -particle.