Question

A proton is traveling to the right at 2.0 * 107 m/s. It has a head on perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton.
What are the speed and direction of each after the collision?

Answer 1

Answer:

(a) $v_{fC} = 3.1\times 10^{- 6}\ m/s$, in the initial direction of the proton.

(b) $v_{fp} - 3.7\times 10^{- 5}\ m/s$, in the opposite direction.

Solution:

As per the question:

Velocity of the proton, $v_{ip} = 2.0\times 10^{7}\ m/s$

If,

Mass of proton  be $m_{p}$

Then

Mass of carbon atom, $m_{C} = 12m_{p}$

Now, if the initial velocity of the proton be $v_{ip}$

Let the carbon atom be initially at rest with initial velocity $v_{iC} = 0\ m/s$

As the collision is perfectly elastic:

Using the principle of conservation of linear momentum:

$m_{p}v_{ip} + m_{c}v_{iC} = m_{p}v_{fp} + m_{c}v_{fC}$

$v_{fp} = 2.0\times 10^{7} - 12v_{fC}$             (1)

Using the conservation of energy:

$\frac{1}{2}m_{p}v_{ip}^{2} + \frac{1}{2}m_{c}v_{iC}^{2} = \frac{1}{2}m_{p}v_{fp}^{2} + \frac{1}{2}m_{c}v_{fC}^{2}$

$\frac{1}{2}m_{p}v_{ip}^{2} + 0 = \frac{1}{2}m_{p}v_{fp}^{2} + \frac{1}{2}12m_{p}v_{fC}^{2}$

Using eqn (1):

$(2.0\times 10^{7})^{2} = (2.0\times 10^{7} - 12v_{fC})^{2} + 12v_{fC}^{2}$

Solving the above quadratic eqn, we get:

$v_{fC} = 3.1\times 10^{- 6}\ m/s$

Since, the velocity here is positive, i.e., the direction of the carbon atom is along the initial direction of proton.

Now, for the speed of the proton:

$v_{fp} = 2.0\times 10^{7} - 12(3.1\times 10^{- 6}) = - 3.7\times 10^{- 5}\ m/s$

Negative sign here indicates that the movement of the proton is in the opposite direction.

Answer 2

Answer:

Vc = 17.83m/s

and Vp = 0.04m/s

This problems was solver by applying the principle of conservation of momentum and energy.

Explanation:

The kinetic energy is conserved.