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
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?
2 answers
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
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