What Physics Concepts Are Applied in These Real-World Scenarios?

[Blake411]

I read the forum rules etc and says you can't do these for me but was wondering if you could suggest what equations/steps to use for them? starting a new chapter and having trouble knowing what to start with, ill do the math myself

1. A hockey puck is moving across the ice at 4.0 m/s. If the coefficient of kenetic friction is 0.030 how far will the puck slide before coming to rest?

2. A pitcher throws a baseball at 30.0 m/s towards a batter. The batter hits the ball back at a velocity of 65.0 m/s. The ball is in contact with the bat for 0.0018 seconds and has a mass of 140 grams.

A) What is the impulse exerted on the ball?

B) What avergae force is exerted on the ball?

3. A ramp is 2.50 meters long and 0.38 meters high. A wood block is slid up the ramp from the bottom at an initial velocity of 4.00 m/s. If the coefficient of kinetic friction is 0.25, what is the velocity of the block at the top of the ramp?

4. An 80.0 kg quarterback is running at 8.0 m/s in the positive "x" direction. He is hit by a 120 kg tackle running at 6.5 m/s at an angle of 40.0 degrees. The two collide and move together, the tackle being careful not to let go of his grip. What is the final velocity and direction of the two?

 

[denverdoc]

What concepts does the chapter cover, obviously impulse and momenta change, what about work?

 

[husky88]

1. The only force acting on the puck is the force of friction, that slows it down.
So Ff=µ*Fn=µ*Fg=µ*m*g
Also F=m*a
So from the above two equations:
µ*m*g=m*a
µ*g=a
From here you can find out the acceleration, which should actually be negative (deceleration).
You know the initial speed (Vo = 4.0 m/s), the final speed (v=0) and the acceleration.
There is a formula that will get you the distance from all this data.

2. a) Impulse is change in momentum. So find out the final momentum (m*v2) and subtract the initial momentum (m*v). The second velocity is negative, as it goes in the opposite direction.
b) F=impulse/t

 

[Blake411]

chapter covers also elastic/inelastic collisions, might have some work combined into it
thanks a bunch huskey for the help, figured out #1 i think check it out=

(coefficient of friction)*g=a
(0.030)*(-9.80 m/s^2) = a
a= -0.294 m/s^2

Vf^2 = Vo^2 +2as
0= (4.0 m/s)^2 + 2(-0.294)s
-16.0 m^2/s^2 = -0.588 m/s^2 (s)
s = 27.2 m

that look alright?

#2-

a)
Pf= P1-P2
Pf= (m*v1) - (m*v2)
Pf= (0.140 kg)(30.0 m/s) - (0.140 kg)(65.0 m/s)
Pf= (4.2 kg*m/s) - (9.1 kg*m/s)
Pf= -4.9 kg*m/s = impulse?

b)
F= impulse/t
F= (-4.9 kg*m/s)/(0.0018s)
F= 2722.2 N

thanks again, help with other 2 problems would be sweet!

 

[husky88]

First one yes. But there are only two significant figures so: 27 m.
Second one, no.
It is final - initial, not initial - final. Also the final has a negative value, as the velocity is in the opposite direction.
You would have -9.1-4.2=-13.3
Or 13.3 in the negative direction. You only have 2 significant figures so the answer would be -13.
The force also has a negative value as it pushes the ball in the negative direction. F= -7400 N (two significant figures).