Combining Momentum with Friction

  • Thread starter student34
  • Start date
  • #1
346
3

Homework Statement



A 1500kg sedan travelling from north to south is struck by a 2200kg truck that is travelling from east to west. The two vehicles become enmeshed due to the impact and slide thereafter. Measurements show the coefficient of friction between the tires and the road is 0.75. The two vehicles slide 5.39m west and 6.43m south of the point of impact. How fast was each vehicle traveling just before the collision?

Homework Equations



P1x = P2x
P1y = P2y
P1x: mx*V2x = V2*(mx + my)
P1y: my*V2y = V2*(mx + my)
P1x = mx*Vt
P1y = V2*(my + mx)
P1y = my*Vy
Ffriction(x) = m(s + t)*a
Ffriction(y) = m(s + t)*a
a = g*μ
V2x = (-2*ax*xskid)^(1/2)
V2y = (-2*ay*yskid)^(1/2)

The Attempt at a Solution



mx*V1x = V2y*(mx + my)
V2y = (2*9.8*0.75*6.43)^(1/2) = 9.722...m/s
V2x = (2*9.8*0.75*5.39)^(1/2) = 8.901...m/s

Now I will substitute V2y and V2x into the conservation of momentum equation:

my*V1y = V2y*(mx + my)

Isolating V1y = (9.722...m/s*3700kg)/1500kg = 23.98...m/s which is wrong because the textbook's answers are 12m/s and 21m/s. I get V1x = 14.97...m/s. Both are wrong by 3m/s, but I have no idea why.
 
Last edited:

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,331
1,012

Homework Statement



A 1500kg sedan travelling from north to south is struck by a 2200kg truck that is travelling from east to west. The two vehicles become enmeshed due to the impact and slide thereafter. Measurements show the coefficient of friction between the tires and the road is 0.75. The two vehicles slide 5.39m west and 6.43m south of the point of impact. How fast was each vehicle traveling just before the collision?

Homework Equations



P1x = P2x
P1y = P2y
P1x: mx*V2x = V2*(mx + my)
P1y: my*V2y = V2*(mx + my)
P1x = mx*Vt
P1y = V2*(my + mx)
P1y = my*Vy
Ffriction(x) = m(s + t)*a
Ffriction(y) = m(s + t)*a
a = g*μ
V2x = (-2*ax*xskid)^(1/2)
V2y = (-2*ay*yskid)^(1/2)

The Attempt at a Solution



mx*V1x = V2y*(mx + my)
V2y = (2*9.8*0.75*6.43)^(1/2) = 9.722...m/s
V2x = (2*9.8*0.75*5.39)^(1/2) = 8.901...m/s

Now I will substitute V2y and V2x into the conservation of momentum equation:

my*V1y = V2y*(mx + my)

Isolating V1y = (9.722...m/s*3700kg)/1500kg = 23.98...m/s which is wrong because the textbook's answers are 12m/s and 21m/s. I get V1x = 14.97...m/s. Both are wrong by 3m/s, but I have no idea why.
You can't break the work (done by friction) into x & y components. Work is a scalar quantity, not a vector.
 
  • #3
346
3
You can't break the work (done by friction) into x & y components. Work is a scalar quantity, not a vector.
I didn't use work or any type of energy equations. I broke acceleration into x and y components.
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,331
1,012
I didn't use work or any type of energy equations. I broke acceleration into x and y components.
The kinematic equation, (vx)2 = 2(ax)x
is equivalent to the work energy theorem (if you include all components).

Since you didn't use work explicitly, let's look at it this way.

For the kinematic equation, (vx)2 = 2(ax)x , you need to use ax, not a. Of course, use ay for to find vy.

The acceleration is in the same direction as the overall displacement. Use that to find its components.
 
  • #5
346
3
The kinematic equation, (vx)2 = 2(ax)x
is equivalent to the work energy theorem (if you include all components).

Since you didn't use work explicitly, let's look at it this way.

For the kinematic equation, (vx)2 = 2(ax)x , you need to use ax, not a. Of course, use ay for to find vy.

The acceleration is in the same direction as the overall displacement. Use that to find its components.
I did that in the work in my original post.
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,838
6,113
P1x: mx*V2x = V2*(mx + my)
P1y: my*V2y = V2*(mx + my)
I can't make sense of that. Do you mean mx*V1x = V2x*(mx + my) etc?
##v_{2x} = (-2 a_x*x_{skid})^{\frac12}##
As SammyS says, this is not valid. A correct form would be
##v^2 = (-2 a*s_{skid})## where ##v^2 = v^2_x+v^2_y; a^2=a_x^2+a_y^2; x_{skid}^2+y_{skid}^2 = s_{skid}^2##, or in vectors (because a and s are collinear) v.v = -2a.s.
 
  • #7
346
3
Wow, I get it, thank you both so much!!!!
 
Last edited:
  • #8
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,331
1,012
I can't make sense of that. Do you mean mx*V1x = V2x*(mx + my) etc?

As SammyS says, this is not valid. A correct form would be
##v^2 = (-2 a*s_{skid})## where ##v^2 = v^2_x+v^2_y; a^2=a_x^2+a_y^2; x_{skid}^2+y_{skid}^2 = s_{skid}^2##, or in vectors (because a and s are collinear) v.v = -2a.s.
That was much clearer than mine.

Thank you haruspex !
 

Related Threads on Combining Momentum with Friction

  • Last Post
Replies
1
Views
2K
Replies
14
Views
193
  • Last Post
Replies
1
Views
29K
Replies
0
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
6
Views
536
  • Last Post
Replies
3
Views
557
Replies
2
Views
2K
Replies
4
Views
6K
Replies
0
Views
2K
Top