The following diagram shows a Ballistic Pendulum. A bullet of mass m is fired on

Question

The following diagram shows a Ballistic Pendulum. A bullet of mass m is fired onto a wooden block of mass M. The bullet embeds itself in the block and the bullet-block system swings upward by distance h as a result of the impact. Answer the questions assuming that m= 0.02kg, M=2.98kg and h=0.4m.

(a) Find the speed of the bullet. (b) What is the speed of the bullet-block system just after the collision? (c Find the mechanical energy loss due to the collision. (d) If the speed is doubled how high will the bullet-block system go?

Explanation / Answer

a)

This is a purely algebraic manipulation. Solving this requires a couple of steps. The first thing to do is figure out a relationship between the initial and final velocities. Since momentum is conserved, we can use that as a starting point:

p0 = mv0
pf = (m+M)v

Since momentum is conserved, we can say that:

p0 = pf

Therefore:

mv0 = (m+M)v

and:

v0 = ((m+M)v / m)

So the only thing left now is figure out an expression for vf. We can do this by analyzing the relationship between kinetic and potential energy:

We know that once point after the bullet hits the ballistic pendulum, the bullet and pendulum mass will fuse and the combined mass will take on kinetic energy. As the mass moves upwards, gravity will slow it down and some of the kinetic energy will be converted to potential energy. When the mass is at it’s maximum height, potential energy will be maximum and kinetic energy will be zero (gravity will have slowed the velocity to zero and the height will be maximized). Vice-versa, when the pendulum is at the bottom (zero height), potential energy will be zero and kinetic energy will be maximum.

We also know that the formula for kinetic energy has velocity in it – but we need a way to remove velocity from our expression. Potential energy doesn’t have velocity in the formula, and we know that at some point, when the pendulum is at a certain height, kinetic energy and potential energy will be equal. So we can set the two energies equal to each other, solve for velocity (in the kinetic energy side), and plug the result into our expression above:

U = (m+M)gh
KE = 1/2(m+M)v2

(m+M)gh = 1/2mv2

v = sqrt(2gh) = sqrt (2*9.8*0.4) = 2.8 m/s

So v0 = ((m+M)v / m) = ((0.02 + 2.98)*2.8) / 0.02 = 420 m/s

Speed of bullet is 420 m/s

b) v = sqrt(2gh) = sqrt (2*9.8*0.4) = 2.8 m/s

c) Mechanical energy is always conserved. so there is no loss

d) if the speed of bullet is doubled,i.e 840 m/s we have to find h

Use equation v0 = ((m+M)v / m)

v = (mv0 / (m+M))

sqrt(2gh) = (mv0 / (m+M))

so, h = (mv0 / (m+M))2 / 2g

h = ((0.02*840)/(0.02+2.98))2 / (2*9.8)

h = 1.6 m

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