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Kathy and Bill collided at an intersection. The police learned the mass of the truck is...

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alejandrogalarce | Student, Undergraduate | (Level 1) Valedictorian

Posted December 18, 2013 at 4:31 AM via web

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Kathy and Bill collided at an intersection. The police learned the mass of the truck is 3200 kg and the mass of the car is 2800 kg. Based on the length of the skid marks at the scene and the mass of the vehicles, police estimate that the combined mass was moving at 7.0 m/s just after impact. From this point, the two vehicles slid and came to rest at the corner of the intersection.

Using the principle of conservation of momentum, the police determined that Bill’s truck was travelling at 2.3 m/s before the collision and Kathy’s car was travelling at 15 m/s before the collision.

The two momentum vectors before the collision are represented by the horizontal and vertical lines below. The momentum after the collision is the slanted line. Adding the two momentum vectors before the collision equals the momentum after the collision, according to the law of conservation of momentum

While the photo suggests a much more complex collision than the given diagram, assume that the diagram shows exactly what happened.

*Magnitude of the momentum after the collision p=m*v=6000kg*7m/s=42000 kg*m/s

*Before the collision the kinetic energy of the car is 315000J

*The kinetic energy of the truck before is 8464J

*The sum of the kinetic energy before the collision is 323464J

*The kinetic energy of the car is after is 68600J

*The kinetic energy of the truck after is 78400J

*The sum of the kinetic energy after the collision is 147000J

1) Using the vector diagram above and your calculated value for the momentum of the combined vehicles, verify the police’s calculations and determine the momentum and velocity of Bill’s truck just before the collision.

2) Using the vector diagram above and your calculated value for the momentum of the combined vehicles, verify the police’s calculations and determine the momentum and velocity of Kathy’s car just before the collision.

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llltkl | College Teacher | (Level 3) Valedictorian

Posted December 18, 2013 at 6:17 AM (Answer #1)

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From triangle ABC in the attached figure,

`tantheta=(BC)/(CA)`

`=p_c/p_t=5.6713`

Again conservation of momentum during the collision requires that,

`p_c^2+p_t^2=(6000*7.00)^2`

`rArr p_t^2(5.6713^2+1)=42000^2`

`rArr p_t^2=42000^2/(5.67^2+1)`

`rArr p_t=7293.22 kg*m/s`

So, `v­_t=(7293.22 kg*m/s)/(3200 kg)`

`=2.28 m/s`

Also

`p_c=5.6713*p_t`

`=41361.93 kg*m/s`

So, `v_c=(41361.93 kg*m/s)/(2800 kg)`

=14.77 m/s

So, the Police estimate of velocities just before the collision is accurate to a fair degree.

a) The momentum and velocity of Bill’s truck just before the collision was 7293.22 kg-m/s and 2.28 m/s respectively.

b) The momentum and velocity of Kathy’s car just before the collision was 41361.93 kg-m/s and 14.77 m/s respectively.

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llltkl | College Teacher | (Level 3) Valedictorian

Posted December 30, 2013 at 4:25 PM (Reply #2)

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You are to assume that the diagram shows exactly what happened.

This shows`theta=mltBAC=(180-100)=80^o` .

So, `tantheta=tan80^o=5.671282~~5.6713`

Again, `P_C/P_T=tantheta=5.6713`

So, `P_C=5.6713*P_T` --- (i)

Conservation of momentum during the collision which creates a triangular scenario requires that,

`P_C^2+P_T^2 =((2800+3200)*7.00)^2`

Make the substitution for `P_C` from (i) to obtain,

`(5.6713*P_T)^2 +P_T^2 =((2800+3200)*7.00)^2`

Take the common term `(P_T^2)` out,

`P_T^2(5.6713^2+1)=42000^2`

`rArr P_T^2=42000^2/(5.6713^2+1)`

Hope that helps.

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alejandrogalarce | Student, Undergraduate | (Level 1) Valedictorian

Posted December 30, 2013 at 9:54 AM (Reply #1)

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Why tan`theta` =5.6713?

I have tan`theta` =5.7065

Why!! Again conservation of momentum during the collision requires that,

`=>` `p^(2)truck` =`(42000)^2/(5.67^2 + 1)` =where the +1 comes from?

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alejandrogalarce | Student, Undergraduate | (Level 1) Valedictorian

Posted January 2, 2014 at 7:52 PM (Reply #3)

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Is the total magnitude of the momentum before the collision the same as the total magnitude of the momentum just after the collision? Explain why or why not.

I don't understand how the momentum is different and what it means:

*momentum before truck is 7293.22 kg m/s

*momentum before car is 41361.93 kg m/s

*Total momentum before is 48655.15 kg m/s

*Total momentum after is 42000 kg m/s

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alejandrogalarce | Student, Undergraduate | (Level 1) Valedictorian

Posted January 2, 2014 at 12:57 AM (Answer #3)

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Is the total magnitude of the momentum before the collision the same as the total magnitude of the momentum just after the collision? Explain why or why not.

I don't understand how the momentum is different and what it means:

momentum before truck is 7293.22 kg m/s

momentum before car is 41361.93 kg m/s

Total momentum before is 48655.15 kg m/s

Total momentum after is 42000 kg m/s 

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