How Do You Calculate the Initial Speed of a Bullet Fired into a Block?

  • Thread starter Thread starter oadeyemi
  • Start date Start date
  • Tags Tags
    Block Bullet
AI Thread Summary
To calculate the initial speed of a bullet fired into a block, the problem involves a bullet of mass m embedding into a block of mass M on a frictionless table. The potential energy of the block after the bullet impacts is given by (m + M)gh, which equates to the kinetic energy ½(m + M)v₂², leading to v₂ = √(2gh). The collision is inelastic, allowing the use of momentum conservation to find the bullet's initial speed, v₁ = ((m + M)/m)v₂. To incorporate the distance d, the time of flight can be calculated using h = 1/2 at², resulting in t = √(2h/g), and thus v₂ can be expressed as d/t. The final formula for the bullet's initial speed becomes v₁ = ((m + M)/m) * [d * √(g/2h)].
oadeyemi
Messages
9
Reaction score
0
1. Homework Statement

A bullet of mass m is fired into a block of mass M initially at rest on a frictionless table of height h. The bullet remains m in the block, and after impact the block lands a distance d from the bottom of the table. Determine the initial speed of the bullet.

2. Homework Equations

I used (m +M)gh for the potential energy and set that equal to the kinetic energy ½(m + M)v₂² and found that v₂ = √(2gh).

And since it was a completely inelastic collision (right?) I used mv₁ = (m + M)v₂ → v₁ = (m + M)v₂ * 1/m

Is this right?


3. The Attempt at a Solution

initial velocity = √(2gh)*((m + M)/m)
 
Physics news on Phys.org
What about the distance d?
 
rl.bhat said:
What about the distance d?

That's the thing, I din't know what do with it. That's why I'm here for help.
 
First solve for the speed of the bullet+block by finding time, (you already have distance):

d=v*t

From kinematic equation h = 1/2 at^2, solve for t: t=\sqrt(2h/g)

Now we know the speed of the bullet+block is just the distance "d" divided by time, so v_2 = d * \sqrt(g/2h).

Using momentum equations, just as you did, v_1 = ((m+M)/m) * v_2 = ((m+M)/m) * [d * \sqrt(g/2h)]
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top