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sineoko [7]
3 years ago
6

Is it ever possible that after an elastic collision (where a moving mass (1) strikes a stationary mass (2)) that the two objects

will have exactly the same final speeds? If so, how must the two masses compare? (Hints, 1st : there are two possibilities as to how the speeds could be equal, 2nd : equations below should be helpful).V1f=V1o (m1-m2/m1+m2) V2f=V1o (2m1/m1+m2)
Mathematics
1 answer:
Veronika [31]3 years ago
4 0

Answer:

Step-by-step explanation:

It is possible that after an elastic collision a moving mass (1) strikes a stationary mass (2) and the two objects will have exactly the same final speed.

During an elastic collision, the momentum and kinetic energy are both conserved.<em> Since one of the object is a stationary object, its velocity will be zero hence the other moving object will collide with the stationary object and cause both of them to move with a common velocity.</em> The expression for their common velocity can be derived using the law of conservation of energy.

<em>Law of conservation of energy states that the sum of momentum of bodies before collision is equal to the sum of momentum of the bodies after  collision.</em>

Since momentum = mass*velocity

<u>Before collision</u>

Momentum of body of mass m1 and velocity u1  = m1u1

Momentum of body of mass m2 and velocity u2  = m2u2

<em>Since the second body is stationary, u2 = 0m/s</em>

Momentum of body of mass m2 and velocity u2  = m1(0) = 0kgm/s

Sum of their momentum before collision = m1u1+0 = m1u1 ... 1

<u>After collision</u>

Momentum of body of mass m1 and velocity vf  = m1vf

Momentum of body of mass m2 and velocity vf  = m2vf

vf is their common velocity.

Sum of their momentum before collision = m1vf+m2vf ... 2

Equating 1 and 2 according to the law;

m1u1 = m1vf+m2vf

m1u1 = (m1+m2)vf

vf = m1u1/m1+m2

<em>vf s their common velocity after collision. This shows that there is possibility that they have the same velocity after collision.</em>

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