The ballistic pendulum is device to measure projectile velocity ...

[aero]

The ballistic pendulum is a simple device to measure projectile velocity v by
observing the maximum angle " to which the box of sand with embedded projectile
swings.
i) Calculate the angle " if the 60-g projectile is fired horizontally into the suspended
20-kg box of sand which is 2 meter in length with a velocity v = 600 m/s.
ii) Find the percentage of energy lost during the impact.

[Melanie]

(i) Let p be the angle formed by the string relative to the vertical when the box achieves its maximum height. The maximum height in this case is L(1 - cos p), where L is the length of the string. Now, if v' is the velocity of the box plus the projectile at the lowest point, we have:

(1/2) (m + M) v'^2 = (m + M)gL(1 - cos p)

by virtue of energy conservation. Here, m is the mass of the projectile, M is the mass of the box. Hence, solving for cos p:

cos p = 1 - v'^2/(2gL)

To solve for v', we use momentum conservation:

mv = (m + M)v'
v ' = mv/(m + M)

Hence:

cos p = 1 - (m/(m + M))^2 v^2/(2gL)

where g = 9.8 m/s^2. Plugging in the values (use SI units!):

cos p = 0.917840634
Ans: p = 23.39 degrees

(ii) The initial energy of the system is:

E = (1/2) mv^2

The final energy of the system is:

E' = (1/2) (m + M) v'^2
E' = (1/2) m^2/(m + M) v^2
E' = (m/(m + M)) E

Thus, the percentage energy lost is:
|E' - E|/E = |(m/(m + M)) - 1|
|E' - E|/E = M/(m + M)
|E' - E|/E = 0.997
Ans: 99.7%

[Jesse_J]

First use conservation of momentum
Momentum before Pb = 0.060*600 = 36kgm/s
Pb = Momentum after Pa
36 = (20+0.06]V => V = 1.795m/s
The kinetic energy of the box with bullet is then
1/2 * 20.06*1.795^2 = 32.3J
This is equal to the potential energy at the peak altitude of the box
= 20.06*9.8*h => h = 0.1643m = 16.43cm
Draw a right triangle with height = 2-0.1643 = 1.8357
and hypotenuse = 2
sin(theta) = 1.8357/2 => theta = 66.6 degrees BUT we need the other angle in this right triangle
= 90-66.6 = 23.39 degrees <---------- i)
which is the angle with the vertical.
KEi = 1/2 * 0.06*600^2 = 10,800J
KEf = 32.3J => Change in KE = 10,768J
10768/10800 = 99.7% of energy lost <-------------- ii)

[TI]

a)  V = Vi*[m/M] = 600*[0.06/20] = 1.8 m/s²
b)  h = V²/2g = 0.1653 m

i)  ? = arccos[1 - h/L] = 23.4°

Ef = E1 - E2 = ½[mVi² - (M+m)V²] = 13.965 J

Ef/Etot = EL/(½mVi²] = 13.965/10800 = 0.00129

%Elost = (1 - (Ef/Etot))*100 = 99.87%