- #1

danago

Gold Member

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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.

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.

_______________________________________________________

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

[tex]

\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}

[/tex]

Therefore, after impact, the velocity of the combined masses is ~7[itex]ms^{-1}[/itex].

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

[tex]

\begin{array}{c}

I = \Delta p \\

= m\Delta v \\

= m(v - u) \\

= 1350({\rm{7}}{\rm{.0557 - 15}}) \\

= - 10724.8{\rm{ }}Ns \\

\end{array}

[/tex]

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[itex]ms^{-1}[/itex] to rest in 5.2m.

[tex]

\begin{array}{l}

v^2 = u^2 + 2as \\

\therefore 0 = 7.0557^2 + 2(5.2)a \\

\therefore a = - 4.79ms^{ - 2} \\

\end{array}

[/tex]

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

[tex]

\begin{array}{c}

\sum F = ma \\

= 2870( - 4.79) \\

= - 13738N \\

\end{array}

[/tex]

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

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

[tex]

\begin{array}{c}

E_M = E_K + E_P \\

= 8m + 294m \\

\end{array}

[/tex]

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:

[tex]

\begin{array}{l}

\frac{{mv^2 }}{2} = 8m + 294m \\

\therefore v^2 = 604 \\

\therefore v = \sqrt {604} = 24.6ms^{ - 1} \\

\end{array}

[/tex]

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:

[tex]

\begin{array}{l}

245m + \frac{{mv^2 }}{2} = 8m + 294m \\

\therefore v = 10.7ms^{ - 1} \\

\end{array}

[/tex]

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

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?

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

**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 [itex]ms^{-1}[/itex]. 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) 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.

[tex]

\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}

[/tex]

Therefore, after impact, the velocity of the combined masses is ~7[itex]ms^{-1}[/itex].

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

[tex]

\begin{array}{c}

I = \Delta p \\

= m\Delta v \\

= m(v - u) \\

= 1350({\rm{7}}{\rm{.0557 - 15}}) \\

= - 10724.8{\rm{ }}Ns \\

\end{array}

[/tex]

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[itex]ms^{-1}[/itex] to rest in 5.2m.

[tex]

\begin{array}{l}

v^2 = u^2 + 2as \\

\therefore 0 = 7.0557^2 + 2(5.2)a \\

\therefore a = - 4.79ms^{ - 2} \\

\end{array}

[/tex]

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

[tex]

\begin{array}{c}

\sum F = ma \\

= 2870( - 4.79) \\

= - 13738N \\

\end{array}

[/tex]

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

_______________________________________________________

**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[itex]ms^{-1}[/itex], 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?

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

a) Assuming that the speed of the cart at point A is 4[itex]ms^{-1}[/itex], 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:

[tex]

\begin{array}{c}

E_M = E_K + E_P \\

= 8m + 294m \\

\end{array}

[/tex]

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:

[tex]

\begin{array}{l}

\frac{{mv^2 }}{2} = 8m + 294m \\

\therefore v^2 = 604 \\

\therefore v = \sqrt {604} = 24.6ms^{ - 1} \\

\end{array}

[/tex]

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:

[tex]

\begin{array}{l}

245m + \frac{{mv^2 }}{2} = 8m + 294m \\

\therefore v = 10.7ms^{ - 1} \\

\end{array}

[/tex]

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

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?

_______________________________________________________

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

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

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.

_______________________________________________________

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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