Ch 7.2 Help

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Ch 7.2 Help

Post  sedwards on Fri Oct 10, 2008 12:50 am

Question 1: (1 point)

NOTE! See the corrected sheet 15' under "Lectures" on the web!

Two cars, car(1) with 1,043 kg mass and a velocity of 15 m/s and car(2) with 2,081 kg mass and unknown velocity approach an intersection at 90o with respect to each other and collide. Immediately after the collision the cars slide stuck together with a velocity of 18.48 m/s. What is the speed of car (2) before the collision (see sheet 10,12-15, 15' summarizes the quick way )?


(How to solve):

Question 2: (1 point)

NOTE! See the corrected sheet 15' under "Lectures" on the web!

Two cars, car(1) with 1,284 kg mass and a velocity of 14.33 m/s and car(2) with 2000 kg mass and unknown velocity approach an intersection at 90o with respect to each other and collide. Immediately after the collision the cars slide stuck together with a velocity of 16.84 m/s. What is the angle between the momentum of the car wreck and the initial direction of car (1) (see sheet 10,12-15, 15' summarizes the quick way)?


(How to solve):


Question 3: (1 point)

Two objects of equal mass m = 2.566 kg collide head-on. Both objects have the velocity of 12.7 m/s. Calculate the total momentum after the collision (see sheet 10,16').


(How to solve):


Question 4: (1 point)

A billiard ball with a kinetic energy 16.26 J makes an elastic collision with another ball which is initially at rest. The collision is such that the two balls have the same kinetic energy after the collision. What is their kinetic energy (see sheet 18)? Indicate with a negative (positive) sign whether the result would be the same (different), if the two balls shared the given kinetic energy before the collision.

(How to solve):

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Ch 7.2 Help

Post  hwilson on Mon Oct 13, 2008 10:32 am

Does anyone know how to solve these?

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Question 1

Post  ravenscorne on Mon Oct 13, 2008 3:33 pm

Question 1: (1 point)

NOTE! See the corrected sheet 15' under "Lectures" on the web!

Two cars, car(1) with 1,043 kg mass and a velocity of 15 m/s and car(2) with 2,081 kg mass and unknown velocity approach an intersection at 90o with respect to each other and collide. Immediately after the collision the cars slide stuck together with a velocity of 18.48 m/s. What is the speed of car (2) before the collision (see sheet 10,12-15, 15' summarizes the quick way )?

(How to solve): (Answer is positive)

Use the equation (m1v1)^2 + (m2v2)^2 = (m3v3)^2 since this is an inelastic collision problem, and it says that the angle is 90 degrees.

So plug everything in and solve for v.

(1043*15)^2 + (2081*v)^2 = (3124*18.48 )^2 Note: m3 = m1 + m2
v = sqrt((3124*18.48 )^2 - (1043*15)^2)/2081
v = sqrt((57731.52)^2 - (15645)^2)/2081
v = 55571.24/2081
v = 26.704 m/s


Last edited by ravenscorne on Mon Oct 13, 2008 3:34 pm; edited 1 time in total (Reason for editing : smilies)

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Question 2

Post  ravenscorne on Mon Oct 13, 2008 3:44 pm

Question 2: (1 point)

NOTE! See the corrected sheet 15' under "Lectures" on the web!

Two cars, car(1) with 1,284 kg mass and a velocity of 14.33 m/s and car(2) with 2000 kg mass and unknown velocity approach an intersection at 90o with respect to each other and collide. Immediately after the collision the cars slide stuck together with a velocity of 16.84 m/s. What is the angle between the momentum of the car wreck and the initial direction of car (1) (see sheet 10,12-15, 15' summarizes the quick way)?

(How to solve): (Answer is Positive)

To solve this do what we did in question one, but afterwards, find the x and y component momentums to find the angle.

v2 = sqrt((m3v3)^2 - (m1v1)^2)/m2
v2 = sqrt((3284*16.84)^2 - (1284*14.33)^2)/2000
v2 = sqrt((55302.56)^2 - (18399.72)^2)/2000
v2 = 52151.93/2000
v2 = 26.076 m/s

θ = tan-1(ΔPx/ΔPy)
θ = tan-1((2000*26.076)/(1284*14.33))
θ = tan-1(52152/18399)
θ = 70.567

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Question 3

Post  ravenscorne on Mon Oct 13, 2008 3:46 pm

Question 3: (1 point)

Two objects of equal mass m = 2.566 kg collide head-on. Both objects have the velocity of 12.7 m/s. Calculate the total momentum after the collision (see sheet 10,16').


(How to solve): (Answer is positive)

This is more of a logic question than anything. Remember that momentum is a vector force, so if two opposing vector forces of the exact same magnitude collide with each other, their resulting magnitude is 0.

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Question 4

Post  ravenscorne on Mon Oct 13, 2008 3:51 pm

Question 4: (1 point)

A billiard ball with a kinetic energy 16.26 J makes an elastic collision with another ball which is initially at rest. The collision is such that the two balls have the same kinetic energy after the collision. What is their kinetic energy (see sheet 1? Indicate with a negative (positive) sign whether the result would be the same (different), if the two balls shared the given kinetic energy before the collision.

(How to solve): (Answer is negative)

For this we use the equation 1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1^2' + 1/2m2v2^2', for the conservation of energy in an elastic collision. But for this equation we can assume the mass of the two balls are the same, so the m drops out of the equation. And since the kinetic energy is equally distributed between the two balls, just divide the given KE by 2.

KE1 = 16.26/2
KE1' and KE2' = 8.13 J

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Re: Ch 7.2 Help

Post  green on Thu Oct 16, 2008 4:59 pm

thanks

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Re: Ch 7.2 Help

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