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?

Answer 1

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

Answer 2

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