Question

A 50 kg mass is travelling at a speed of 2 m/s. Another 60 kg
mass travelling at a speed of 12 m/s in the same direction,
strikes the first mass. After the collision the 50 kg mass is
travelling with a speed of 4 m/s. The coefficient of restitution
of the collision is
a) 19/30
b)30/19
c)20/11
d)11/20​

Answer 1

Answer:

Coefficient of restitution for the collision is 19/30

Explanation:

As we know that there is no external force on the system of two mass

So here we can use momentum conservation for the collision

So we will have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

now we have

$50(2) + 60(12) = 50(4) + 60v_{2f}$

$100 + 720 = 200 + 60 v$

$620 = 60 v$

$v_{2f} = \frac{31}{3}$

now by formula of coefficient of restituition we will have

$e = \frac{v_{2f} - v_{1f}}{v_{1i} - v_{2i}}$

here we will have

$e = \frac{\frac{31}{3} - 4}{2 - 12}$

$e = \frac{\frac{19}{3}}{-10}$

$e = \frac{19}{30}$

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Answer 2

The coefficient of restitution of the collision=$\frac{19}{30}$

Explanation:

$m_1=50 kg$

$m_2=60 kg$

$v_{1i}=2 m/s$

$v_{2i}=12 m/s$

$v_{1f}=4 m/s$

We know that

Linear momentum is conserved in the given collision

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

Substitute the values then we get

$50\times 2+60\times 12=50\times 4+60v_{2f}$

$820=200+60v_{2f}$

$60v_{2f}=820-200=620$

$v_{2f}=\frac{620}{60}$

$v_{2f}=\frac{31}{3} m/s$

Coefficient of restitution

$e=\frac{v_{2f}-v_{1f}}{v_{2i}-v_{1i}}$

Substitute the values then we get

Coefficient of restitution

$e=\frac{\frac{31}{3}-4}{12-2}=\frac{19}{30}$

Hence, the coefficient of restitution of the collision=$\frac{19}{30}$

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