Question

A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at60^ofrom the original direction. After the collision, what are the speeds of (a) the target proton and (b) the projectile proton?

Short answer

a) Speed of target proton after collision is 433m/s

b) Speed of projectile proton after collision 250 m/s

Step-by-step solution

Listing the given quantities

Speed of target proton before collision is $v_{tx}=0\mathrm{m}/\mathrm{s}$.

Speed of projectile proton before collision is .$v_{px}=500\mathrm{m}/\mathrm{s}$

Angle made by projectile proton after collision with +x axis is, $\theta =60°$

Angle between target proton and projectile proton after collision is$90°$

Understanding the concept of law of conservation of momentum

Applying the law of conservation of momentum in x and y direction to the given system we can get two equations. Solving those two and inserting given values we can find the speed of target proton and the projectile proton after collision.

Formula:

$$\begin{gathered} P_{ix}=P_{fx} \\ {P}_{iy}=P_{fy} \end{gathered}$$

(a) Calculation of speed of target proton after collision

Let the mass of proton be m.

Since the angle made by the projectile proton after collision with +x axis is $\theta =60°$and the angle between the target proton and projectile proton after collision is $90°$,the angle made by the target proton after collision with +x axis is,$\alpha =\left(60-90\right)°=-30°$

According to the law of conservation of momentum,

$$\begin{gathered} P_{i,x}=P_{f,x} \\ m{v}_{px}+m{v}_{tx}=m{v}_p^{\text{'}}\cos \theta +m{v}_t^{\text{'}}\cos \alpha \\ 500=0.5v_p^{\text{'}}+0.866v_t^{\text{'}}.....\left(1\right) \\ {P}_{i,y}=P_{f,y} \end{gathered}$$

$$\begin{gathered} m{v}_{ty}+m{v}_{py}=m{v}_p^{\text{'}}\sin \theta +m{v}_t^{\text{'}}\sin \alpha \\ 0=v_p^{\text{'}}\sin \theta +v_t^{\text{'}}\sin \alpha \\ 0=v_p^{\text{'}}\sin \left(60°\right)+v_t^{\text{'}}\sin \left(-30°\right) \\ 0=0.866v_p^{\text{'}}-0.5v_t^{\text{'}}......\left(2\right) \end{gathered}$$

Solving equation (1) and (2) we get,

$v_t^{\text{'}}=433\mathrm{m}/\mathrm{s}$

Therefore, the speed of the target proton after collision is 433 m/s

(b) Calculation of speed of target proton after collision 

From part (a),

$$\begin{gathered} 0=0.866v_p^{\text{'}}-0.5v_t^{\text{'}} \\ 0.866v_p^{\text{'}}=0.5v_t^{\text{'}} \\ {v}_t^{\text{'}}=\frac{0.5}{0.866}\left(433\right) \\ =250\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, speed of the projectile proton after collision is 250 m/s