# Few Problems

Gold Member
Hey. I've got a physics test coming up soon, and i had a few questions that I am wanting confirmed. All help is greatly appreciated.

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1. IN a horrific car crash, a car skids 85m before striking the rear of a parked car. Just before the moment of impact, the car is moving with a velocity of 15 $ms^{-1}$. The cars become locked together and skid for another 5.2m before coming to rest. The mass of the first car is 1350kg, and the mass of the second is 1520kg.

a) What is the velocity of the two cars just after impact?

b) What is the impulse on each car during the collision?

c) What is the average size of the frictional forces between the cars and the road that finally bring the cars to rest?

a) For part a, i just used the fact that momentum is conserved.

$$\begin{array}{l} \sum {p_i } = 1350(15) = 20250 \\ \sum {p_f } = (1350 + 1520)v = 2870v \\ \sum {p_i } = \sum {p_f } \\ \therefore 2870v = 20250 \\ \therefore v = 7.0557ms^{ - 1} \\ \end{array}$$

Therefore, after impact, the velocity of the combined masses is ~7$ms^{-1}$.

b) Part b, i just found the change in momentum.

$$\begin{array}{c} I = \Delta p \\ = m\Delta v \\ = m(v - u) \\ = 1350({\rm{7}}{\rm{.0557 - 15}}) \\ = - 10724.8{\rm{ }}Ns \\ \end{array}$$

Therefore, the impulse experienced by each car is 10724.8 Ns.

c) I had the most trouble with part c. For this question, i found the acceleration of the cars, from ~7$ms^{-1}$ to rest in 5.2m.

$$\begin{array}{l} v^2 = u^2 + 2as \\ \therefore 0 = 7.0557^2 + 2(5.2)a \\ \therefore a = - 4.79ms^{ - 2} \\ \end{array}$$

Therefore, the masses were deccelerating at a rate of 4.79$ms^{-2}$. For this decceleration to occur in a mass of 2870kg, i used Newtons formula to calculate the force required:

$$\begin{array}{c} \sum F = ma \\ = 2870( - 4.79) \\ = - 13738N \\ \end{array}$$

Therefore, an average force of ~13738N is required to bring the two cars to rest within 5.2m of collision.

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2. The following image depicts a roller coaster:

http://img148.imageshack.us/img148/6906/pfq0002copyst1.gif [Broken]

a) Assuming that the speed of the cart at point A is 4$ms^{-1}$, find the speed of the cart at points B, C and D. Also assume that no energy is lost due to friction/drag.

b) It is found that at point C, the cart stops. What is the energy loss due to friction between points A and C?

I was a little confused with this one at first. I started by finding the total mechanical energy at point A:

$$\begin{array}{c} E_M = E_K + E_P \\ = 8m + 294m \\ \end{array}$$

Assuming that this is a perfectly elastic energy transformation, the mechanical energy at point B should be the same as the kinetic energy, since potential energy is zero, because of a zero height relative to the ground. So:

$$\begin{array}{l} \frac{{mv^2 }}{2} = 8m + 294m \\ \therefore v^2 = 604 \\ \therefore v = \sqrt {604} = 24.6ms^{ - 1} \\ \end{array}$$

At point C, since energy is still being conserved, and this is an elastic situation, the mechanical energy should equal the mechanical energy used in the previous calculation. So that:

$$\begin{array}{l} 245m + \frac{{mv^2 }}{2} = 8m + 294m \\ \therefore v = 10.7ms^{ - 1} \\ \end{array}$$

And i used the same process for point D to get a velocity of 19.2$ms^{-1}$.

For question b, I am quite stuck. I am not sure where to start from. Any help please? Would the answer just be the kinetic energy the cart has at point C, which would be ~57m J? Also, how can i calculate the efficiency of the track from A to C?

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3. A ball of mass 0.1kg strikes a raquet with velocity of 10$ms^{-1}$, and rebounds with a velocity of 9.5$ms^{-1}$. Using the force/displacement graph, find the maximum displacement of the strings.

http://img225.imageshack.us/img225/3824/scancopybz1.gif [Broken]

This question I am completely stuck. Everything I've tried leads nowhere. My physics teacher couldn't even do it.

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Thanks very much for the help. For the questions I've answered, id just like some confirmation as to weather I've done them correctly.

Thanks again,
Dan.

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Gold Member
Ok. Sorry about accidently posing before finishing my post. Its all done now though :)

J77
May help

For part 3 - the slope of the force-displacement curve gives you the stiffness coefficient.

Gold Member
Stiffness coefficient? Thats something that I've never come across or heard of. Where can i find more information on this?

Staff Emeritus
Gold Member
Well presented. I don't have tim to check them all immediately but I've checked the first. All the methods are correct and the numbers seem fine nice work.

Homework Helper
#3. Many assumptions need to be made in order to get to an answer for this one. I think the basic idea is that the raquet acts like a spring storing the kinetic energy of the ball temporarily and then returning it back to the ball (like an object being dropped onto a spring, but in this case only kinetic energy needs to be considered). It seems there is some energy lost in this interaction, but I would assume that the strings would deflect according to the initial kinetic energy of the ball. Which means that one need to find the displacement at which the area under the graph (the work done by the strings) is the same as the initial kinetic energy of the ball.

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wheres point A on the rollercoaster?

Gold Member
wheres point A on the rollercoaster?

Oops accidently cut it off. Its the point 30m from the ground :)

And thanks for the replies everyone. I am still not quite sure about question 3 though

J77
The work done in part 3 is $$\frac12 K x^2$$ where K's the stiffness coefficient.

Can this be equated to anything?

Gold Member
ok. So the stiffness coefficient is 200. Do i then just find the maximum value of x in:

$$W = 100x^2$$

?

Gold Member
Nvm. Just realized the quadratic has no maximum.

Staff Emeritus
danago said:
Stiffness coefficient? Thats something that I've never come across or heard of. Where can i find more information on this?
Stiffness is the propensity to resist displacement when subject to a force. The greater the stiffness, the higher the spring constant, i.e. the more force is takes for a given displacement. This is the meaning of Hooke's law.

http://en.wikipedia.org/wiki/Hooke's_law - relates force and displacement
http://en.wikipedia.org/wiki/Young's_modulus - relates stress and strain (which are related to stress and strain)

The spring constant is an indication of the spring's stiffness. A large value for indicates that the spring is stiff. A low value for means the spring is soft.
http://phoenix.phys.clemson.edu/labs/124/shm/index.html

Gold Member
Thanks for that astro.

Staff Emeritus
Along the lines of what J77 is alluding, what is the origin of the energy in the spring (tennis racket springs) when it deflects and what is the velocity of the ball at maximum spring deflection?

Gold Member
The energy to displace the strings would be the kinetic energy from the ball, and at maximum displacement, the balls velocity is zero?

Staff Emeritus
At maximum displacement, the spring has stopped moving (deflecting) - then it starts moving in the opposite direction. That's more or less the definition of an extremum.

Gold Member
Im still not really seeing how to solve the problem :(

Staff Emeritus
Looking back at post #6 (andrevdh) and post #9 (J77), one could do an equivalence between the spring energy 1/2 kx2 and the initial kinetic energy of the ball 1/2 m v2. Now, as andrevdh implied, requires the assumption that the raquet plane does not deflect itself, i.e. all the deflection goes into the springs.

The above relationship simplies to kx2 = m v2, which is where J77 was leading you.

Now the other part is the fact that the ball rebounds with a velocity of 9.5 m/s, so there is a loss of energy somewhere, which could be internal friction due to inelastic collision (deformation of the ball). One could assume half the energy loss during the extension of the spring and the other half during restoration, so then the effective kinetic energy going into the strings may be based on the mean velocity (speed) of 9.75 m/s.

Conservation of energy applies to elastic collision, and not inelastic collision.

J77
I was thinking of equating the work done by the 'spring' with the change in kinetic energy of the ball...

Would this not work?

Staff Emeritus
Gold Member
For #2 part b. you know the speed it should be going if there is no friction, so due to the friction all that speed has diminished, therefore the energy loss is as you thought. Efficiency is normally defined as the real performance/ideal performance. I suggest using energy in such a manner

efficiency= energy lost/initial energy

seems most logical to me. I could be wrong because I've never had to apply it to this situation though so if somebody could check me.

Staff Emeritus
J77 said:
I was thinking of equating the work done by the 'spring' with the change in kinetic energy of the ball...

Would this not work?
The change in kinetic energy would be dissipated or transformed somewhere.

Initially the ball has kinetic energy based on v= +10 m/s. At maximum deflection (extension), the ball comes to rest, v = 0. Then the spring begins to restore or return to its equilibrium, but when the ball leaves the spring, it has lesser velocity, -9.5 m/s (speed = 9.5 m/s in opposite direction), where the - simply indicates that the velocity is in the opposite direction.

As andrevdh, there are some assumptions that have to be made. For example, is the ball rigid or non-rigid. The ball may undergo elastic or inelastic deformation, which means it could dissipate some of the energy during the interaction with the racquet. Or the racquet itself could recoil.

One must assume how the energy (difference in kinetic energies) is dissipated. A reasonable assumption is that half is dissipated during the initial deflection and half dissipated during the restoration.

This is another reason why it is very helpful to work out a problem as danago has done.

J77
OK - I see where you're coming from but should one make such assumptions when asked for the maximum displacement or when doing physics tests?

Staff Emeritus