Block slides across a sled on frictionless ice....

[Yoonique]

Homework Statement

Two students sit on opposite ends of a sled 6 m long initially at rest on frictionless ice. Each student has a mass of 50 kg; the sled’s mass is 40 kg. The student at one end slides a 2-kg object on the sled across to the other at a uniform speed of 5 m/s relative to the sled (the object moves friction free on the sled). (a) What is the sled’s speed relative to the ice before and after the second student catches the object? (b) Over what distance does the sled move while the object slides across? (c) Over what distance does the centre of mass move while the object slides across?

Homework Equations

ΣP_i = ΣP_f
x_cm = (1/Σm_i) Σ(m_ix_i)
x-horizontal right = positive

The Attempt at a Solution

a) Sled’s speed relative to the ice:
ΣP_i = ΣP_f
0 = 2.0(5) - 40v ( v = velocity of sled with respect to inertial frame)
v= -0.25
v_s/i = -0.25 -5 = -5.25
After: 0m/s

b) Taking the center of the sled as (0,0)
x_cm = [2.0(-3)+50(-3)+50(3)+40(0)] / (2.0+50+50+40) = -0.0423m
x_cm = [2.0(3) + 40d] / (142) = -0.0423m
d = -0.3m
which means the sled moves 0.3m to the left

c) Since ΣP_i = 0, ΣP_f = CM x v = 0
thus CM does not move at all so d = 0

Is my answers correct?

 

[AlephNumbers]

0 = 2.0(5) - 40v ( v = velocity of sled with respect to inertial frame)
v= -0.25
vs/i = -0.25 -5 = -5.25

Aren't the students on the sled? Would they not also move with the sled?

 

[Yoonique]

Aren't the students on the sled? Would they not also move with the sled?

So the students will have the same velocity as the moving sled with respect to the ground?
0 = 2.0(5) - 40v - 50v - 50v
v= -0.0714
so the velocity of sled with respect to ice is -0.0714-5= -5.0714

 

[AlephNumbers]

so the velocity of sled with respect to ice is -0.0714-5= -5.0714

That would be the velocity of the sled with respect to the object. They ask for the velocity of the sled with respect to the ice.
The velocity of the object with respect to the ice would be 5+V_sled. Try substituting that into

0 = 2.0(5) - 40v - 50v - 50v

I'm sorry, that should be 5+V_sled

 

[AlephNumbers]

Actually, the way you wrote the momentum it should be 5-V_sled because you are not treating v as a vector quantity.

 

[Yoonique]

That would be the velocity of the sled with respect to the object. They ask for the velocity of the sled with respect to the ice.
The velocity of the object with respect to the ice would be 5+V_sled. Try substituting that into

I'm sorry, that should be 5+V_sled

Oh, my mistake. So the velocity of sled with respect to ice should be -0.0714.
Is part c correct?

 

[AlephNumbers]

Oh, my mistake. So the velocity of sled with respect to ice should be -0.0714.

That is not correct. Like I said, go back to ∑P_i = ∑P_f, but consider the velocity of the object with respect to the ice. You are given the velocity of the object with respect to the sled, 5 m/s. To find the velocity of the sled with respect to the ice with ∑P_i = ∑P_f, you must use the velocity of the object with respect to the ice, which is v_o/i = v_o/s + v_s/i.

 

[Yoonique]

That is not correct. Like I said, go back to ∑P_i = ∑P_f, but consider the velocity of the object with respect to the ice. You are given the velocity of the object with respect to the sled, 5 m/s. To find the velocity of the sled with respect to the ice with ∑P_i = ∑P_f, you must use the velocity of the object with respect to the ice, which is v_o/i = v_o/s + v_s/i.

So I will get v_o/i = 5 + 0 = 5 + v_s/i
∑Pi = ∑Pf
0 = 2.0(5+v_s/i) + 140v_s/i where I treat v as a vector quantity
v_s/i = -0.0704
When writing equation for conservation of momentum, is it better to treat v as a vector quantity or include - sign and treat is as a scalar quantity?

 

[AlephNumbers]

So I will get vo/i = 5 + 0 = 5 + vs/i
∑Pi = ∑Pf
0 = 2.0(5+vs/i) + 140vs/i where I treat v as a vector quantity
vs/i = -0.0704

Beautiful.

I find it easier to treat the velocities as vector quantities and sum them. It is basically the same either way, just make sure to be consistent.

 

[AlephNumbers]

Also, your solution to b is incorrect. You have the velocity of the sled relative to the ice now, so try to find the amount of time over which the motion occurs. From there it should be pretty simple to find the distance the sled travels.

 

[AlephNumbers]

But c looks good.

 

[Yoonique]

Also, your solution to b is incorrect. You have the velocity of the sled relative to the ice now, so try to find the amount of time over which the motion occurs. From there it should be pretty simple to find the distance the sled travels.

d=0.0704 x 6/5 = 0.0845m
Since the centre of mass of the system will not move, can I use the formula of centre of mass to find the distance moved by the centre of mass of the sled when the object reach the other end of the sled? I assume you can't because the sled moves so the coordinate system changes and we do not know the new coordinates now?

 

[AlephNumbers]

You can find the center of mass of the sled after the object has slid to the opposite end of the sled using the relevant equation that you listed. This is because you know that the motion will stop after the object reaches the other side of the sled, and the sled will have traveled that distance d from its original position. I would take the original position of the center of mass of the sled to be the origin. The new coordinates of the center of mass of the sled are then x = -d.

 

[Yoonique]

You can find the center of mass of the sled after the object has slid to the opposite end of the sled using the relevant equation that you listed. This is because you know that the motion will stop after the object reaches the other side of the sled, and the sled will have traveled that distance d from its original position. I would take the original position of the center of mass of the sled to be the origin. The new coordinates of the center of mass of the sled are then x = -d.

Thanks, both methods managed to get the same answer for d.

 

[S Chanda]

if the block does not move with uniform velocity and stops at the other end of the sled due to friction ,then what will be the final velocity of the sled when the block stops?