Question #132672
a ball mass 2kgg is mvoing with velocity 5m/sec elastically collide with a ball of mass 5 kg which is moving with velocity 2m/s find the speed of separation.
1
Expert's answer
2020-09-14T10:22:22-0400

Total momentum a total energy of the system is conserved in the elastic collisions. We can write following:

m1v1+m2v2=m1v1+m2v2m_1 v_1+m_2 v_2 = m_1 v_1'+m_2 v_2'

m1v12+m2v22=m1v12+m2v22m_1v_1^2 + m_2 v_2^2 = m_1v_1'^2 + m_2v_2'^2 (here we have already divided full equation by 2)


m1(v1v1)=m2(v2v2)m_1(v_1 - v_1')=m_2(v_2'-v_2) (1)

m1(v1+v1)(v1v1)=m2(v2+v2)(v2v2)m_1(v_1+v_1')(v_1-v_1')= m_2(v_2' +v_2)(v_2'-v_2) (2)

dividing (1) by (2):

v1+v1=v2+v2v_1 +v_1' = v_2'+v_2 (3)


After solving (1)+(2) and taking into account (3) we have:

v1=v2+2m1v1+m2v2m1+m2=2+210+107=3.714v_1' = -v_2 + 2 \frac{m_1v_1+m_2v_2}{m_1+m_2} = -2+ 2 \frac{10+10}{7} = 3.714 m/s.

v2=v1+2m1v1+m2v2m1+m2=5+210+107=0.714v_2'= -v_1+ 2 \frac{m_1v_1+m_2v_2}{m_1+m_2} = -5 + 2 \frac{10+10}{7}=0.714 m/s.


After the collision the balls are moving into opposite directions, so the speed of separation is

v1+v2=3.714+0.714=4.428v_1'+v_2'= 3.714 + 0.714 = 4.428 m/s.

Answer: 4.4284.428 m/s.



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