Question

the body of mass 10kg moving with a velocity of 5 m/s hits a body of 1gm at rest. The velocity of the second body after collision, assuming it to be perfectly elastic is (in m/s)​

Answer 1

Answer:

ANSWER

For a perfectly elastic collision, coefficient of restitution is e=1.

Let u and v be the velocities of 10 kg and 1 g masses after collision.

Thus, e=

rel. vel. of approach

rel. vel.of separation

=

5−0

v−u

=

5

v−u

But, as e=1, we have v−u=5.

Total momentum before collision is P=10×5=50 kg m/s

Total momentum after collision is P=10×(v−5)+0.001×v=10.001v−50

Thus, 10.001v−50=50⇒v=

10.001

100

≈10 m/s

Explanation:

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Answer 2

Answer:

Velocity of the second body after collision = 10 m/s

Explanation:

Given:

• Mass of the first body = 10 kg
• Mass of the second body = 1 gm = 0.001 kg
• Initial velocity of the first body =  5 m/s
• Initial velocity of the second body = 0 m/s
• The collision is perfectly elastic

To Find:

• Final velocity of the second body

Solution:

Since this is a one dimensional elastic collision we use the formula,

$\tt v_2=\dfrac{m_2-m_1}{m_1+m_2}\times u_2+\dfrac{2m_1}{m_1+m_2}\times u_1$

where v₂ = final velocity of second body

m₂ = mass of second body

m₁ = mass of first body

u₂ = initial velocity of second body

u₁ = initial velocity of first body

Substituting the data we get,

$\tt v_2=\dfrac{0.001-10}{10+0.001} \times0+\dfrac{2\times 10}{10+0.001}\times 5$

$\tt v_2=0+\dfrac{20}{10.001} \times 5$

$\tt v_2 =\dfrac{100}{10.001}$

$\tt v_2=9.99\:m/s \approx 10\:m/s$

Hence the velocity of the second body after collision is 10 m/s.

Notes:

To find the velocity of the first body after collision in an elastic one dimensional collision we use the formula,

$\tt v_1=\dfrac{m_1-m_2}{m_1+m_2}\times u_1+\dfrac{2m_2}{m_1+m_2} \times u_2$