Question

36. A proton of mass 1.67 x 10 kg collides elastically
head-on with an a-particle of mass 6.68 x 10-2 kg,
initially at rest. After the collision, the a-particle
moves with a speed of 8 x 10 ms. Find the speed of
the proton before and after the collision.​

Answer 1

Answer:

$\displaystyle{v_1=0.637 \ m/sec}\\\\\displaystyle{v_2=-79.36 \ m/sec}$

Explanation:

Given ;

Two masses 1.67 x 10 kg and  6.68 x 10⁻² kg

The by conservation of momentum ;

$\displatystyle{m_1u_1+m_2u_2=m_1v_1+m_2v_2}\\\\\displatystyle{Here \ u_2=o$

$\displaystyle{m_1u_1=m_1v_1+m_2v_2 \ ..(i )}\\\\\displaystyle{For \ elastically \ e=1}\\\\\displaystyle{1=\dfrac{v_2-v_1}{u_1u_2}}\\\\\displaystyle{u_1-u_2=v_2-v_1 \ ..(ii)}$

From ( i )  and ( ii )  we get

$\displaystyle{v_1=\dfrac{ 2m_2u_2}{m_1+m_2}}\\\\\\\displaystyle{v_2=\dfrac{(m_2-m_1)u_2}{m_1+m_2}$

putting given value in formula we get

$\displaystyle{v_1=\dfrac{ 2\times6.68\times10^{-2}\times8\times10}{6.68\times10^{-2}+1.67\times10}}\\\\\\ \displaystyle{v_1=0.637 \ m/sec}\\\\\\\displaystyle{v_2=\dfrac{(6.68\times10^{-2}-1.67\times10)8\times10}{6.68\times10^{-2}+1.67\times10}}\\\\\\\displaystyle{v_2=-79.36 \ m/sec}$

Thus we get answer .

Answer 2

Answer :-

Explanation :-

Given :-

mass of proton (m1) = 1.67 × 10 kg

mass of particle (m2) = 6.68 × 10^-2 kg

speed of particle before collision (u2) = 0 m/s

speed of particle after collision (v2) = 8 × 10 m/s

_________________________

To find :-

I) speed of proton before collision (u1).

II) speed of proton after collision (v1).

_________________________

Solution :-

Using Law of Conservation of linear momentum,

m1u1 + m2u2 = m1v1 + m2v2

Since,