Question

A head-on, elastic collision between two particles with equal initial speed v leaves the more massive particle (mass m1) at rest. find the ratio of the particle masses

Answer 1

1/3 The key thing to remember about an elastic collision is that it preserves both momentum and kinetic energy. For this problem I will assume the more massive particle has a mass of 1 and that the initial velocities are 1 and -1. The ratio of the masses will be represented by the less massive particle and will have the value "r" The equation for kinetic energy is E = 1/2MV^2. So the energy for the system prior to collision is 0.5r(-1)^2 + 0.5(1)^2 = 0.5r + 0.5 The energy after the collision is 0.5rv^2 Setting the two equations equal to each other 0.5r + 0.5 = 0.5rv^2 r + 1 = rv^2 (r + 1)/r = v^2 sqrt((r + 1)/r) = v The momentum prior to collision is -1r + 1 Momentum after collision is rv Setting the equations equal to each other rv = -1r + 1 rv +1r = 1 r(v+1) = 1 Now we have 2 equations with 2 unknowns. sqrt((r + 1)/r) = v r(v+1) = 1 Substitute the value v in the 2nd equation with sqrt((r+1)/r) and solve for r. r(sqrt((r + 1)/r)+1) = 1 r*sqrt((r + 1)/r) + r = 1 r*sqrt(1+1/r) + r = 1 r*sqrt(1+1/r) = 1 - r r^2*(1+1/r) = 1 - 2r + r^2 r^2 + r = 1 - 2r + r^2 r = 1 - 2r 3r = 1 r = 1/3 So the less massive particle is 1/3 the mass of the more massive particle.

Answer 2

The ratio of the particle masses is \boxed{\frac{1}{3}} or \boxed3 .

Further explain:

We have to calculate the ratio of the particle masses.

As we know, in the elastic collision between two masses the momentum and the energy both are conserved.

Here, the collision between the masses the head-on it means head to head.

For head on head collision the masses will travel parallel but opposite in the direction.

We have two masses one is heavier and another is lighter.

The mass of massive or heavier particle is ${m_1}$.  

The mass of the lighter particle is ${m_2}$.  

From the conservation of linear momentum total initial momentum is equal to the total final momentum.

Therefore,

$\boxed{\left( {{m_1}v - {m_2}v} \right) = \left( {{m_1}{v_1} + {m_2}{v_2}} \right)}$

Here, after the collision the massive particle comes into rest.

So, final expression will be,

$\left( {{m_1}-{m_2}}\right)v={m_2}{v_2}$                                   …… (1)

From the conservation of the energy,

Total kinetic energy before collision is equal to the total kinetic energy after collision.

Therefore,

$\begin{aligned}\frac{1}{2}{m_1}{v^2}+\frac{1}{2}{m_2}{v^2}&=\frac{1}{2}{m_2}{\left( {{v_2}} \right)^2}\\{m_1}{v^2}+{m_2}{v^2}&={m_2}{\left( {{v_2}}\right)^2}\\\left( {{m_1}+{m_2}}\right){v^2}&={m_2}{\left( {{v_2}}\right)^2}\\\end{aligned}$ 

Simplify the above equation,

$\begin{aligned}{m_2}{\left( {{v_2}} \right)^2}&=\frac{{\left( {{m_1}+{m_2}} \right){v^2}}}{{{m_2}}}\\{v_2}&=\left( {\sqrt {\frac{{\left( {{m_1}+{m_2}} \right)}}{{{m_2}}}} }\right)v\\\end{aligned}$

 

Substitute the value of ${v_2}$ in equation (1).

$\begin{aligned}\left( {{m_1} - {m_2}} \right)v&={m_2}\left( {\sqrt {\frac{{\left( {{m_1} + {m_2}}\right)}}{{{m_2}}}} } \right)v \\\left( {{m_1} - {m_2}} \right)&=\sqrt {{m_2}\left( {{m_1} + {m_2}}\right)}\\{m_2}\left( {\frac{{{m_1}}}{{{m_2}}} - 1}\right)&={m_2}\sqrt {\left( {\frac{{{m_1}}}{{{m_2}}} + 1} \right)}\\\left( {\frac{{{m_1}}}{{{m_2}}}-1}\right)&=\sqrt {\left( {\frac{{{m_1}}}{{{m_2}}}+ 1}\right)}\\\end{aligned}$

 

Substitute $x$ for$\dfrac{{{m_1}}}{{{m_2}}}$ in above equation.

$\left( {x - 1} \right)=\sqrt {\left( {x + 1} \right)}$

 

Squaring both the sides in above equation,

$\begin{aligned}{\left( {x - 1} \right)^2}&=\left( {x + 1}\right)\\{x^2} - 2x + 1&=x + 1\\{x^2}-3x&=0\\\end{aligned}$

 

Taking $x$ as a common in the above equation.

$x\left( {x - 3} \right)=0$

On solving above equation

We get,

$x = 3$

Replace the value of $x$  

$\boxed{\frac{{{m_1}}}{{{m_2}}} = 3}$

 

Or,

$\boxed{\frac{{{m_2}}}{{{m_1}}} = \frac{1}{3}}$  

Learn more:

1. Average kinetic energy: brainly.com/question/9078768

2. Broadcast wavelength of the radio station: brainly.com/question/9527365

3. Motion under force brainly.com/question/7031524.

Answer details:

Grade: Senior School

Subject: Physics

Chapter: Impulse and Momentum

Keywords:

Head on collision, two particles, equal speed, ratio of particle masses, momentum, conservation of momentum, energy, conservation of energy, masses, ratio.