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Q28PE

Expert-verifiedFound in: Page 290

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Two identical objects (such as billiard balls) have a one-dimensional collision in which one is initially motionless. After the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic energy are conserved.**

Both the momentum and kinetic energies are conserved.

**The product of a particle's mass and velocity is called momentum. The term "momentum" refers to a quantity that has both a magnitude and a direction. The temporal rate of change in momentum is equal to the force applied on the particle, according to Isaac Newton's second law of motion.**

Mass of the first ball (\({m_1}\)) = Mass of the second ball (\({m_2}\))

Speed of the first ball before the collision, \({v_1}\) = Speed of the second ball after the collision, *v*_{2}′

Speed of second ball before collision, \({v_2}\) = Speed of first ball after collision, *v*_{1}′ = 0 m/s

The Law of conservation of momentum states that the total momentum of a system always remains constant before and after collisions or we can say that the initial momentum before the collision of a system is equal to the final momentum of the system after the collision for an isolated system.

Let \({p_1}\)_{ }and \({p_2}\)_{ }be the initial momentum of two objects before the collision, \(p_1^\prime \)and \(p_2^\prime \) (primed variables represent values after collision) be the final momentum after the collision, then according to the law of conservation of momentum,

\({p_1} + {p_2} = p_1^\prime + p_2^\prime \)..........(1)

In this problem, after the collision, the two balls do not stick together and move with different velocities. These types of collisions are known as elastic collisions.

Substituting the value of \(1{p = mv}\) in equation 1, we get

\({m_1}{v_1} + {m_2}{v_2} = m_1^{}v_1^\prime + {m_2}v_2^\prime \).......(2)_{ }

where \(v_1^\prime \) and \(v_2^\prime \) are the final velocities of the balls 1 and 2 after collision respectively.

Put the given values in equation 2, gives

\(\begin{align}{}{m_1}{v_1} + {m_2} \times \left( {0m/s} \right) &= m_1^{} \times (0m/s) + {m_2}v_2^\prime \\{m_1}{v_1} &= {m_2}v_2^\prime \end{align}\) ----------- (3)

Since \({m_1} = {m_2}\), then \({v_1} 2 = v {_2^\prime 2}\)

Thus from equation 3, momentum before the collision is equal to the momentum after the collision, so we can say that in the case of collisions between two billiard balls momentum is conserved.

The Law of conservation of kinetic energies states that the total kinetic energy of a system always remains constant before and after collisions or we can say that the initial kinetic energy before the collision of a system is equal to the final kinetic energy of the system after the collision.

We have the equation for the kinetic energy of a particle of mass *m, *\(K.E = \dfrac{1}{2}m{v^2}\)

Thus according to the law of conservation of energy, \(K.E = K.E\prime \) .................(4)

* *Thus equation 4 becomes,* *

\(\begin{aligned}\dfrac{1}{2}{m_1}{v_1} 2 + \dfrac{1}{2}{m_2}{(0m/s) 2} &= \dfrac{1}{2}{m_1}{(0m/s) 2}\dfrac{1}{2}m_2^{}v{_2 \prime 2}\\\dfrac{1}{2}{m_1}{v_1}^2 = \dfrac{1}{2}m_2^{}v{_2 \prime ^2}\end{aligned}\)

Since \({m_1} = {m_2}\), then \({v_1} 2 = v{_2^\prime 2}\)

So, \({v_1} = v_2^\prime \)

Thus kinetic energy before the collision is equal to the kinetic energy after the collision indicates that the kinetic energy is also conserved.

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