here is my question: A bullet with a mass of 6.00 g is fired through a 1.25 kg block of wood on a frictionless surface. The initial speed of the bullet is 896 m/s, and the speed of the bullet after it exits the block is 435 m/s. At what speed does the block move after the bullet through it.
here is my answer: mass of bullet = 6gm
= 6/1000 kg = .800kg
let 'v' be speed of block. apply law of conservation of energy
initial energy of (block+bullet)
= final energy of (block+bullet)
kinetic energy=1/2 mV^2 (formula)
solve for v: 1/2x.006*(896)^2=1/2.006*(435)^2 +1/2(1.25)(v)^2
1.25v^2=.006(896^2-435^2)
v=54.27m/s
please show me if there is a simpler way to write this out. please write out the equaiton and fill in the appropriate data.
i think what i have is too confusing
No. Energy is not converved when the bullet hits the block, momentum, however, is.
Momentumblock=changemomentumbullet
1250*vblock=6*(896-435)
solve for vblock
Idk fam
can you please show me?
Certainly! Here's a simpler way to write out the equation and fill in the appropriate data:
Let's denote the mass of the bullet as m_bullet, the mass of the block as m_block, the initial speed of the bullet as v_initial_bullet, the final speed of the bullet as v_final_bullet, and the speed of the block as v_block.
We can use the law of conservation of momentum to solve for v_block:
m_bullet * v_initial_bullet + m_block * 0 = m_bullet * v_final_bullet + m_block * v_block
Since there is no friction or external forces acting on the system, the momentum of the bullet-block system remains constant.
Now, let's substitute the given values into the equation:
m_bullet = 6.00 g = 0.006 kg
m_block = 1.25 kg
v_initial_bullet = 896 m/s
v_final_bullet = 435 m/s
0.006 kg * 896 m/s + 1.25 kg * 0 = 0.006 kg * 435 m/s + 1.25 kg * v_block
0.005376 kg·m/s = 0.00261 kg·m/s + 1.25 kg * v_block
Now, isolate v_block:
0.005376 kg·m/s - 0.00261 kg·m/s = 1.25 kg * v_block
0.002766 kg·m/s = 1.25 kg * v_block
v_block = 0.002766 kg·m/s ÷ 1.25 kg
v_block ≈ 0.0022128 m/s
So, the speed at which the block moves after the bullet passes through it is approximately 0.0022 m/s.