A ball of mass 0.25 kg is fired with velocity 120 m/s into the barrel of a sprin

Question

  A ball of mass 0.25  kg is fired with velocity 120  m/s into the barrel of a spring gun of mass 2.1  kg initially at rest on a frictionless surface. The ball sticks in the barrel at the point of maximum compression of the spring. No energy is lost to friction. What fraction of the ball's initial kinetic energy is stored in the spring?

A ball of mass 0.25 kg is fired with velocity 120 m/s into the barrel of a spring gun of mass 2.1 kg initially at rest on a frictionless surface. The ball sticks in the barrel at the point of maximum compression of the spring. No energy is lost to friction. What fraction of the ball's initial kinetic energy is stored in the spring?

Explanation / Answer

we have to use conservation of energy, and linear momentum. We can see that the energy the ball had is 1/2 mv^2 or kinetic energy. Using 0.5 mv^2 formula, we can see K.E = 0.5 *0.25*120^2 = 1800J.

Now, after the ball hits the spring gun, they start moving in same velocity. As no external force was applied, momentum remains conserved.

i,e,   initial m*v = final m*v.

Initial m*v = 0.25 * 120.

final m * v = (2.1+0.25) * x where x is the velocity they end up with(combined gun and ball)

or, 0.25 * 120 = 2.35 * x

so x = 12.8 m/s

So this is the final speed. So final kinetic energy = 1/2 mv^2 = 0.5 * (2.1+0.25) * 12.8^2 = 192.5 J.

We started with 1800J and ended up with 192.5 J. So 1800 - 192.5 = 1607 J must have been stored in the spring. i.e about 89.27% or 0.892 fraction of the the ball's kinetic energy is now in the loaded spring.

hope this helps

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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