(a) Prove that the ratio of the kinetic energy of the projectile–pendulum system

Question

(a) Prove that the ratio of the kinetic energy of the projectile–pendulum system immediately after the collision to the kinetic energy immediately before is m1/(m1 + m2).

(b) What is the ratio of the momentum of the system immediately after the collision to the momentum immediately before?

1

(m1 + m2)/m1


m1/m2

m1/(m1 + m2)

m1^2/(m1^2 + m2^2)

(c) A student believes that such a large decrease in mechanical energy must be accompanied by at least a small decrease in momentum. How would you convince this student of the truth?

Explanation / Answer

Let the mass of the projectile be m1

the mass of the pendulum be m2

then the momentum before collision is pi = m1v

the momentum after collision pf = (m1 + m2) vf

from law of conservation of momentum

       m1v = (m1 + m2) vf

then vf = m1v / (m1 + m2)

Now the energy before collision is

           Ki = 1/2 m1 v^2

the energy after collision

          Kf = 1/2 (m1 + m2) vf^2

therefore the ratio

        Kf/Ki = [(1/2)(m1 + m2) vf^2] / (1/2) m1v^2

                 = [(m1 + m2) (m1/(m1+m2))^2 v^2] / m1 v^2

                  = (m1) /(m1 + m2)

(b) The ratio of momenta

       pf/pi = (m1 + m2)vf / m1v

                = (m1 + m2)[ m1v/(m1 + m2)] / m1v

                = 1

(c) The relation between the energy and momentum

             K = p^2/2m

The kinetic energy is directly proportional to the square of the momentum

so when there is samll decrease in momentum changes the energy square of its

therefore there is large change in energy

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.