bullet of mass 15 gm is shot horizontally into a 15 kg block of wood attached to a spring of spring constant 1000 N/m. The bullet remains inside the block after impact. The force of the impact compresses the spring 10 cm. What is the velocity of the bullet?
You have a conversation of EP and with that you have the formula:
Mbull*Vi+Mwood*Vi=(Mb+Mw)Vf
.o15kg*Vi+15*0=(.015+15)Vf
Mult. by 0:
.015kg*Vi=(.o15+15)Vf
So we can't solve that but we move on to where the spring compresses, there is conservation of TME so:
1/2(Mb+Mw)Vf^2=(1/2)kx^2
1/2(.015+15)Vf^2=(1/2)1000*(.1m)^2
7.5075*Vf^2 = 5
------ -----
7.5075 7.5075
than take the square root of that number:
Vf=.816
Plug it into the first unfinished formula:
.015kg*Vi=(.o15+15)Vf
.015kg*Vi=(.o15+15)(.816)
.015*Vi=.1836
---- -----
.015 .015
Vi=12.24 m/s
You have a conversation of EP and with that you have the formula:
Mbull*Vi+Mwood*Vi=(Mb+Mw)Vf
.o15kg*Vi+15*0=(.015+15)Vf
Mult. by 0:
.015kg*Vi=(.o15+15)Vf
So we can't solve that (i guess we cant curse here) but we move on to where the spring compresses, there is conservation of TME so:
1/2(Mb+Mw)Vf^2=(1/2)kx^2
1/2(.015+15)Vf^2=(1/2)1000*(.1m)^2
7.5075*Vf^2 = 5
------ ----- (divide by 7.5075)
7.5075 7.5075
than take the square root of that number:
Vf=.816
Plug it into the first unfinished formula:
.015kg*Vi=(.o15+15)Vf
.015kg*Vi=(.o15+15)(.816)
.015*Vi=.1836
---- ----- (divide by .015)
.015 .015
Vi=12.24 m/s
To find the velocity of the bullet, we can use the principle of conservation of momentum. The momentum before the impact is equal to the momentum after the impact.
The momentum before the impact can be calculated by multiplying the mass of the bullet (15 gm = 0.015 kg) by its velocity (which we want to find), so let's call it v_b:
Momentum before = 0.015 kg * v_b
The momentum after the impact is equal to the combined momentum of the bullet and the block. Since the bullet remains inside the block after the impact, their velocities will be the same.
The mass of the block is given as 15 kg, and the velocity of the block and bullet combined after the impact is zero, since the block comes to rest at maximum compression.
Momentum after = (15 kg + 0.015 kg) * 0
Using the conservation of momentum principle, we can set these two equations equal to each other:
0.015 kg * v_b = (15 kg + 0.015 kg) * 0
Simplifying the equation, we find:
0.015 kg * v_b = 0
This implies that the velocity of the bullet, v_b, is 0 m/s. This means that the bullet comes to a complete stop when it hits the block.
Please note that in real-world situations, there would be some energy loss due to friction and other factors, so the bullet might not come to a complete stop. However, in this hypothetical scenario, where no energy loss is taken into account, the bullet's velocity is zero after impact.