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1 End of Chapter Test

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.
Which parts do you need help with?
Reply 2
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ollyeland
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Original post by GabiAbi84
Which parts do you need help with?

question 2 onwards lol
Original post by ollyeland
question 2 onwards lol

Post what you've tried ...
Original post by ollyeland
question 2 onwards lol


Okay so let’s start with question2
What have you done so far?
Reply 5
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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
Reply 6
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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
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?
Reply 8
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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
Original post by ollyeland
J = Δp = p₂ - p₁ = mv₂ - mv₁ = mΔv


Okay so what’s your figures for v1 and v2?
Reply 10
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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
Reply 11
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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?
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
(edited 3 years ago)
Reply 13
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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
Reply 14
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ollyeland
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Original post by ollyeland
yeh thats what i got

i got the answer, its -4800
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
(edited 3 years ago)
Reply 16
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ollyeland
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ok how
Reply 17
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ollyeland
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is it 17ms
Original post by ollyeland
is it 17ms


That’s it! 👍🏻
Reply 19
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ollyeland
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so the final answer for 2 is -9600

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