Question

In: Physics

A spring gun fires a bullet of mass m=.04 kg horizontally at a ballistic pendulum apparatus...

A spring gun fires a bullet of mass m=.04 kg horizontally at a ballistic pendulum apparatus with a mass M=.350. the bullet lodges itself into the pendulum. After collision, the center of mass of the bullet and pendulum rises by .07 meters. What is the approximate initial speed v of the bullet?

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Expert Solution

In ballistic pendulum experiment:

A bullet of mass 'm' traveling with speed 'v' strikes with a pendulum of mass 'M', then this pendulum rises to a height of 'h' due to collision.

Now Since after collision projectile + pendulum system rises to height 'h', So using energy conservation:

KEi + PEi = KEf + PEf

PEi = 0, at reference point

KEf = 0, Speed of system zero at height 'h'

KEi = (1/2)*(m + M)*V^2

PEf = (m + M)*g*h

So,

(1/2)*(m + M)*V^2 + 0  = 0 + (m + M)*g*h

V = sqrt (2*g*h)

h = rise in center of mass of pendulum = 0.07 m

V = sqrt (2*9.81*0.07) = 1.172 m/s

So now using momentum conservation in this case, speed of pendulum before collision will be

Pi = Pf

m*v + M*0 = (m + M)*V

v = (m + M)*V/m

Using given values: U = Initial speed of bullet = ? m/s

m = mass of bullet = 0.04 kg

M = mass of pendulum = 0.350 kg, So

v = (0.350 + 0.04)*1.172/0.04

v = 11.43 m/s

Let me know if you've any query.


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