The Student Room Group

M1 Momentum and Impulse Question

Hi, I'm doing the May 2002 M1 paper. We only studied momentum and impulse briefly before Easter so please forgive any blatant incompetence!

"The masses of two particles A and B are 0.5kg and mkg respectively. The particles are moving on a smooth horizontal table in opposite directions and collide directly. Immediately before the collision the speed of A is 5m/s and the speed of B is 3m/s. In the collision, the magnitude of the impulse exerted by B on A is 3.6 Ns. As a result of the collision the direction of the motion of A is reversed.

(a) Find the speed of A immediately after the collision.

The speed of b immediately after collision is 1m/s.

(b) Find the two possible values of m."

OK, part (a) seems simple enough...

I = m(v-u)
3.6 = 0.5(v-5)
7.2 = v-5
v = 12.2m/s

I assume I'm meant to use total initial momentum = total final momentum for part (b)...

2.5 + 3m = 6.1 + m
2m = 3.6
m = 1.8kg

How do I get two values??? I'm slightly unsure about "the direction of motion of A is reversed" does that just mean 12.2m/s should be -12.2m/s? Even so it doesn't seem affect my answer for part (b)?

I'm sure I missing something obvious...any help would be very much appreciated.
nooneknows
Hi, I'm doing the May 2002 M1 paper. We only studied momentum and impulse briefly before Easter so please forgive any blatant incompetence!

"The masses of two particles A and B are 0.5kg and mkg respectively. The particles are moving on a smooth horizontal table in opposite directions and collide directly. Immediately before the collision the speed of A is 5m/s and the speed of B is 3m/s. In the collision, the magnitude of the impulse exerted by B on A is 3.6 Ns. As a result of the collision the direction of the motion of A is reversed.

(a) Find the speed of A immediately after the collision.

The speed of b immediately after collision is 1m/s.

(b) Find the two possible values of m."

OK, part (a) seems simple enough...

I = m(v-u)
3.6 = 0.5(v-5)
7.2 = v-5
v = 12.2mf

ok assume a has negative direction intially
then 3.6=0.5(v-(-5))
v=2.2ms^-1
this might be wrong but 12.2 would seem rather large

then to find the values of m
inital momentum is (0.5*-5)+(m*3)
=-2.5+3m
final momentum = (2.2*0.5)+(m*1)
=11+m
2m=8.5
m=4.25
does that help in some way? im not too sure about this as i would prefer to write it down than do it in my head. The would be two values as we have shown depending on which way you had a and b going intially. that doesnt seem to make sense but its the only thing i can think of and would happen because of the fact that impulse effects the magnitude and direction of the velocity.
Reply 3
a) remember, impulse on B = -impulse on A

-3.6 = 0.5(v-5) => v = -2.2 m/s
diablo2004
a) remember, impulse on B = -impulse on A

-3.6 = 0.5(v-5) => v = -2.2 m/s

of course i thought the value for velocity that noonesknow has was a little too large. do you know why there are two possible values for m tho?
Reply 5
Speciez99
of course i thought the value for velocity that noonesknow has was a little too large. do you know why there are two possible values for m tho?

The 2 values of m come from "the speed of B after the collision is 1m/s", you have to consider v=1 and v=-1 and substitue these into 3.6=m(3+v) giving m = 0.9 or 1.8 kg
Reply 6
Impulse is a vector so has a direction.
I = m(v-u)
-3.6 = 0.5(v-5)
-7.2 = v-5
v = -2.2m/s

We are given thast the speed is 1 m/s but not told in what direction!

So, the momentum eqn becomes,

0.5*5 + m*(-3) = 0.5*(-2.2) + m*(±1)
2.5 - 3m = -1.1 ± m
3.6 = (3 ± 1)m
m=3.6/4 or m=3.6/2
m=0.9 kg or m=1.8 kb
Reply 7
Aaargh, too late :frown:
Reply 8
Fermat
Aaargh, too late :frown:

Yep, in future don't act like your namesake and keep your solutions/proofs to yourself :tongue:
Reply 9
Bezza
Yep, in future don't act like your namesake and keep your solutions/proofs to yourself :tongue:

:biggrin: :biggrin:
Reply 10
Wow! Unbelievably fast responses! Thank you.

Yeh, I thought that 12.2 was probably a bit large, but I wasn't sure how to get anything different....everything makes sense now though. Thanks again!