How far do the runner and catcher slide?

[btrkun]

Homework Statement

It's the bottom of the ninth inning at a baseball game. Score is tied with a runner on 2nd when the batter gets a hit. The 85 kg runner rounds 3rd and is heading home with a speed of 8.0 m/s. Just before he reaches home plate he crashes into the catcher and the two players slide together along the base path toward home plate. Catcher's mass is 95kg. Coefficient of friction is 0.70. How far does the runner and catcher slide?

Homework Equations

m1iv1i + m2iv2i = m1fv1f + m2fv2f

The Attempt at a Solution

I found the velocity at which they slide together to be:

m1iv1i +m2 = (m1 +m2)vf
(85 kg)(8.0 m/s) + (95 kg) = (180 kg)vf
680 kg*m/s + 95 kg = 180 kg*vf
vf = 4.3 m/s

So now I have mass = 85+95 = 180 kg; vi = 4.3 m/s; vf = 0; u = 0.70

I don't know what to do next to find the displacement. The two equations I know to find x are:
x=1/2(vf+vi)delta-t
x=(vi)(delta-t)+1/2a(delta-t)^2

But I don't know t. I don't know a since I don't know t, unless I'm suppose to assume its constant. even with a=0 i still don't know t. please help!

 

[yellowgators]

you may want to go back first to where you substituted values into your original equation. what is the velocity of the catcher when the runner collides with him? if it's zero, it will cancel out the mass of the catcher entirely from the first part of the equation.
also, maybe there's an equation you can use to substitute known values such as Vf and Vi for t...

 

[Doc Al]

I found the velocity at which they slide together to be:

m1iv1i +m2 = (m1 +m2)vf
(85 kg)(8.0 m/s) + (95 kg) = (180 kg)vf
680 kg*m/s + 95 kg = 180 kg*vf
vf = 4.3 m/s

Careful: What's the initial speed of the catcher? (Correct this and recalculate vf.)

So now I have mass = 85+95 = 180 kg; vi = 4.3 m/s; vf = 0; u = 0.70

I don't know what to do next to find the displacement. The two equations I know to find x are:
x=1/2(vf+vi)delta-t
x=(vi)(delta-t)+1/2a(delta-t)^2

You can use another kinematic formula that directly relates distance with speed, without needing the time. (Or you can just combine these two equations to eliminate t.)

But I don't know t. I don't know a since I don't know t, unless I'm suppose to assume its constant. even with a=0 i still don't know t.

You don't need the time to calculate the acceleration. Analyze the forces. (That's why they give you the coefficient of friction.)

 

[btrkun]

I see my error. The vf = 308 m/s

I'm still kinda lost on the equation bit. I was thinking F=ma

Fy = mg-N = 0
Fx = ma -uN ??

 

[Doc Al]

I see my error. The vf = 308 m/s

Way off: Redo it more carefully.

I'm still kinda lost on the equation bit. I was thinking F=ma

Fy = mg-N = 0

Good.

Fx = ma -uN ??

What's the only horizontal force acting on the players and thus the net horizontal force? Set that equal to ma and solve for a.

 

[btrkun]

Honestly, I don't know why I typed 308 m/s... I meant to type 3.8 m/s.

The only force acting horizontally is u = 0.70

So it's:
0.70 = ma - N ??

If so then how do I find N? mgcos(0) = (85)(9.8)cos(0) = 833?

 

[Doc Al]

Honestly, I don't know why I typed 308 m/s... I meant to type 3.8 m/s.

OK. (But better not to round off until the last step.)

The only force acting horizontally is u = 0.70

The only force acting horizontally is friction, which equals $\mu N$.

So it's:
0.70 = ma - N ??

??

If so then how do I find N? mgcos(0) = (85)(9.8)cos(0) = 833?

Use the entire mass of both players. You find N by analyzing vertical forces, like you did in post #4: N-mg = 0, so N = mg.

 

[btrkun]

Ahhhhhh, okay.uN = ma
(0.70)[(180 kg)(9.8 m/s^2)] = (180 kg)a
a = 6.86 m/s^2

knowing a, I can figure out t to be
t = a/v
t = (6.86 m/s^2) / (3.7 m/s)
t = 1.85 s

x = vt
x = (3.7 m/s)(1.85 s)
x = 6.84 m
x = 7mBetter?

 

[Doc Al]

uN = ma
(0.70)[(180 kg)(9.8 m/s^2)] = (180 kg)a
a = 6.86 m/s^2

Good.

knowing a, I can figure out t to be
t = a/v
t = (6.86 m/s^2) / (3.7 m/s)
t = 1.85 s

[Edit] Not good: Correct that formula; it should be t = v/a. And be careful about rounding off.

x = vt
x = (3.7 m/s)(1.85 s)
x = 6.84 m
x = 7m

Careful! The speed is not constant, so you'd better use the average speed to find the distance. And you'll need to correct the time.

 

[btrkun]

bleh, this problem has turned into the bane of my existence...

So average speed is v = (delta-x) / (delta-t)
To find x I use delta-v?

x = (delta-v)(delta-t)
x = (vf - vi)(tf - ti)
x = (0 -3.8 m/s)(1.8 s - 0) (t = (6.86 m/s^2)(3.8 m/s))
x = -6.84 m

It's the same answer, just negative. How can distance be negative?

 

[tiale11]

what do you guys think of this?
I am going to solve the problem using energy
u=Ff/Fn
0.7*180kg*9.81m/s^2=Ff
Ff=1236.06N

1/2 m*v^2 = Ff *d
1/2 180kg*(3.8m/s)^2= 1236.06N *d
d = 1.04m

is that correct Doc Al?

 

[Doc Al]

knowing a, I can figure out t to be
t = a/v
t = (6.86 m/s^2) / (3.7 m/s)
t = 1.85 s

I must have been asleep before. You have the formula switched around. It should be t = v/a.

 

[Doc Al]

bleh, this problem has turned into the bane of my existence...

So average speed is v = (delta-x) / (delta-t)
To find x I use delta-v?

x = (delta-v)(delta-t)

The equation needed is x = (average speed) x (time), not (change in speed) x (time).

The two equations I know to find x are:
x=1/2(vf+vi)delta-t

That's the one you want.

And don't forget to redo your calculation of the time, as pointed out in posts #9 and 12.