- #1
Juntao
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A cart of mass m = 6 kg carrying a spring of spring constant k = 48 N/m and moving at speed v = 1.5 m/s hits a stationary cart of mass M = 12 kg. Assume all motion is along a line.
a) What is the speed of the center of mass of this system?
b) When the spring is at its maximum compression, with what speed are the carts moving in the lab frame?
c) How far will the spring be compressed?
d) After the collision is over, what is the velocity of m in the center-of-mass frame? (right is positive)
e) After the collision is over, what is the velocity of m in the lab frame?
f) After the collision is over, what is the velocity of M in the lab frame?
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Ok, I figured out a, and b, but now I'm stuck on c.
This is what I've done so far.
a) Initial momentum=final momentum
(m1v1+m2v2)=(m1+m2)Vcom
where m1=6kg
v1=1.5m/s
m2=12kg
v2=0
Vcom=velocity of center of mass
I figured out the answer of Vcom=.5m/s
b) I know that the answer is .5 m/s, but I'm sure why it is though. Can someone elaborate on this please?
c) This is where I get stuck so far. I used the idea of conservation of momentum and energy. In the end, I get this equation
KE=SPE or
.5(m1+m2)(Vcom)^2=.5kx^2
So I substituted my numbers, and the x I get isn't right. Is this the right way of apporaching it?
a) What is the speed of the center of mass of this system?
b) When the spring is at its maximum compression, with what speed are the carts moving in the lab frame?
c) How far will the spring be compressed?
d) After the collision is over, what is the velocity of m in the center-of-mass frame? (right is positive)
e) After the collision is over, what is the velocity of m in the lab frame?
f) After the collision is over, what is the velocity of M in the lab frame?
======================================================
Ok, I figured out a, and b, but now I'm stuck on c.
This is what I've done so far.
a) Initial momentum=final momentum
(m1v1+m2v2)=(m1+m2)Vcom
where m1=6kg
v1=1.5m/s
m2=12kg
v2=0
Vcom=velocity of center of mass
I figured out the answer of Vcom=.5m/s
b) I know that the answer is .5 m/s, but I'm sure why it is though. Can someone elaborate on this please?
c) This is where I get stuck so far. I used the idea of conservation of momentum and energy. In the end, I get this equation
KE=SPE or
.5(m1+m2)(Vcom)^2=.5kx^2
So I substituted my numbers, and the x I get isn't right. Is this the right way of apporaching it?