ollyeland
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I need help with this Physics End of Chapter Test. Thanks

Two cars are involved in a head-on collision.
a State two quantities conserved in the collision. [2]
b The collision between the cars is described as an inelastic collision.
Explain what is meant by an inelastic collision. [2]
2 A car of mass 1200 kg is travelling along a straight road at a velocity of 25 m s−1
.

The driver applies the brakes for a duration of 2.0 s.
During this interval, the car decelerates with a magnitude of 4.0 m s−2
.

Calculate the change in the momentum of the car. [3]
3 A 2.8-tonne lorry moving at a speed of 30 m s−1

collides into the back of a stationary car

of mass 800 kg on the hard shoulder of the motorway.
During the collision, the lorry and the car get tangled together.
a Calculate the common speed V of the tangled lorry and car. [4]
b Calculate the loss of kinetic energy in the collision. [3]
4 A bumper car of mass 300 kg moving at 4.0 m s−1

collides with another bumper car of mass

420 kg moving at 1.5 m s−1 in the opposite direction.
After the collision, the speed of the 300 kg car decreases to 0.50 m s−1

but it carries on moving

in the same direction (see diagram).

Determine the speed v and the direction of the 420 kg bumper car after the collision.
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GabiAbi84
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Which parts do you need help with?
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ollyeland
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(Original post by GabiAbi84)
Which parts do you need help with?
question 2 onwards lol
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Muttley79
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(Original post by ollyeland)
question 2 onwards lol
Post what you've tried ...
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GabiAbi84
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(Original post by ollyeland)
question 2 onwards lol
Okay so let’s start with question2
What have you done so far?
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ollyeland
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(Original post by GabiAbi84)
Okay so let’s start with question2
What have you done so far?
well at first I thought the answer was 25200, but im not sure
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ollyeland
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(Original post by Muttley79)
Post what you've tried ...
I've not been ablw to do anything
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GabiAbi84
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(Original post by ollyeland)
well at first I thought the answer was 25200, but im not sure
Post your working for it.
Where did you get that number from?
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ollyeland
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(Original post by GabiAbi84)
Post your working for it.
Where did you get that number from?
J = Δp = p₂ - p₁ = mv₂ - mv₁ = mΔv
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GabiAbi84
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(Original post by ollyeland)
J = Δp = p₂ - p₁ = mv₂ - mv₁ = mΔv
Okay so what’s your figures for v1 and v2?
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ollyeland
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(Original post by GabiAbi84)
Okay so what’s your figures for v1 and v2?
v1=25 v2=21 because decellerates by 4
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ollyeland
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(Original post by GabiAbi84)
Okay so what’s your figures for v1 and v2?
v1=25 v2=21 because decellerates by 4ms. Is the equation right?
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GabiAbi84
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(Original post by ollyeland)
v1=25 v2=21 because decellerates by 4ms. Is the equation right?
The deceleration is at 4ms^-2 over the t=2s

So a= -4
Last edited by GabiAbi84; 1 month ago
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ollyeland
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(Original post by GabiAbi84)
The deceleration is at 4ms^-2 over the t=2s

So a= -4
yeh thats what i got
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ollyeland
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(Original post by ollyeland)
yeh thats what i got
i got the answer, its -4800
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GabiAbi84
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(Original post by ollyeland)
yeh thats what i got
No , you have said that it decelerates by 4ms which is incorrect.

You need to work out v2 using suvat
Last edited by GabiAbi84; 1 month ago
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ollyeland
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ok how
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ollyeland
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is it 17ms
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GabiAbi84
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(Original post by ollyeland)
is it 17ms
That’s it! 👍🏻
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ollyeland
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so the final answer for 2 is -9600
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