Ballistic Pedulum - Finding Kinetic Energy Lost

AI Thread Summary
The discussion focuses on solving a problem involving a ballistic pendulum used to measure the speed of a projectile. The initial velocity of the bullet is calculated using conservation of momentum, leading to the equations for kinetic energy before and after the collision. The kinetic energy lost during the collision is determined by subtracting the final kinetic energy of the pendulum and bullet system from the initial kinetic energy of the bullet. The calculated kinetic energy lost amounts to 0.74J, indicating the energy dissipated during the collision. This analysis aids in understanding the bullet's speed and the collision's efficiency.
ScoutFCM
Messages
17
Reaction score
0
Here's a problem that I've been having trouble on for awhile and seem to be stuck. I was just wondering if someone could guide me or show me how to do this problem.

The Ballistic pendulum is a device usd to measure the speed of a fast-moving projectile such as a bullet. The bullet is fired into a large block of wood suspended from some light wires. The bullet is stopped by the block, and the entire system swings through the vertical distance, h. The mass of the bullet (m1=0.068kg), the mass of the pendulum (m2=0.256kg), h=6.2cm.

Vo=(.324kg/.068kg) x (2 x 9.8 m/s^2 x .062m)^1/2 = 5.25m/s

KEinitial= 1/2(.068kg x 5.25m/s)^2 = 0.937J

KEfinal= 1/2(.324kg) x (2 x 9.8 m/s^2 x .062m) = .197J

I got that far. I was wondering how do I find the kinetic energy lost from the info?
 
Physics news on Phys.org
If I'm understanding your problem correctly, you're trying to solve for vo, the initial velocity of the bullet and the bullet/pendulum comes to a rest when it swings up to a height h.

Let's call the mass of the bullet m and the mass of the block M and let's set our datum for potential energy at the initial height of the bullet/pendulum.

First we need to consider the collision of the block and bullet using the conservation of momentum. This will give us the initial velocity of the ballistic pendulum:

m*vo = (m+M)v

Since we're neglecting things like air resistance and friction at the pendulum pivot, we know that all of this kinetic energy will be converted into potential energy:

.5*(M+m)v^2 = (M+m)*g*h

So find the expression for v in terms of vo, then plug into the 2nd equation to solve for vo.

If you need the kinetic energy lost in the collision, you can calculate the kinetic energy before and after:

KEb = .5*m*vo^2

KEa = .5*(M+m)*v^2

and find the difference.
 


To find the kinetic energy lost in this problem, we can use the conservation of energy principle. This states that the total energy in a closed system remains constant, meaning that the initial kinetic energy of the bullet (KEinitial) is equal to the final kinetic energy of the combined system (KEfinal). Therefore, the kinetic energy lost can be calculated by subtracting the final kinetic energy from the initial kinetic energy.

In this case, the kinetic energy lost would be:

KElost = KEinitial - KEfinal

Substituting the values we have calculated, we get:

KElost = 0.937J - 0.197J = 0.74J

Therefore, the kinetic energy lost in this scenario is 0.74J. This means that the bullet lost 0.74J of its initial kinetic energy when it collided with the pendulum. This information can be useful in determining the speed of the bullet before it hit the pendulum, as well as other factors such as the efficiency of the collision and the amount of work done by the pendulum. I hope this helps guide you in solving the problem.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top