M2 question Review Exercise 2

ok its review exercise 2 , p151 in M2.

I insist that the answer to part a) is wrong/mis-print, unless I've done something seriously wrong. The answer I get to a) is 4/3(root (ga)) and not simply root(ga).

please help.

Reply 1

It'sPhil...

2

~Revol~

ok its review exercise 2 , p151 in M2.

I insist that the answer to part a) is wrong/mis-print, unless I've done something seriously wrong. The answer I get to a) is 4/3(root (ga)) and not simply root(ga).

please help.

Its possible. What is the question and we could have a look to see whose answer we agree with

Reply 2

~Revol~

OP

Qu. A light inextensible string of length 4a has a particle A, of mass m, attached at one end and a particle B, of mass m, attached at the other end. The string passes through a small smooth ring which is fixed at a point O at a distance 3a above a horizontal table. The system is hanging in equilibrium with OB=2a when a smooth bead of mass 2m, which is threaded on the string between O and B, is released from rest at O. The bead falls under gravity until it collides with and adheres to the particle B to form a composite particle C.

a) Given that the string remains taut, show that the speed of C immediately after the collision is (ga)^0.5

Reply 3

~Revol~

OP

omg, I cant believe how irritating M2 questions in review exercise 2 are! They seem so wrong!, I havent managed to do one of them from (Qu20 onwards) properly yet.

Also someone try these too, and tell me what answers u get! coz im pretty sure they are dodgy (either the answers or the question itself....)

Qu.21) A cricket match is played on level ground. Just before the batsman hits the ball, it is 1m above the ground and travelling with speed 20m/s in a direction inclined at 30 degrees below the horizontal. Immediately after the ball has been hit it has speed 25 m/s and the direction of motion is reversed. The ball has mass 0.15kg.

The horizontal distance from the batsman to A (the point where the ball first reaches the ground), is 56.9m.

Find the speed of the ball when it reaches A.

Qu 24) This is ONE WEIRD F****** QUESTION! How the **** do you draw the diagram to start with???? :

A smooth horizontal rail is fixed at a height 3h above a rough horizontal ground. A uniform rod AB, of mass M and length 6h, is placed in a vertical plane perpendicular to the rail with the end A resting on the ground. The distance AC = 5h, where C is the point of contact between the rail and the rod. Show that the force exerted by the rail on the rod is of magnitude 12Mg/25.

CAN U ACTUALLY DRAW A DIAGRAM FOR THIS THAT MAKES RELATIVE SENSE WITHOUT?

Reply 4

Hoofbeat

14

That review exercise is horrible, particualrly the moments questions - my mates and I all gave up.

Re: Q24 do a search for a thread by me asking for help with M2/Moments question as someone very kindly drew a diagram of how that is supposed to look :smile:

Reply 5

baba

~Revol~

Qu. A light inextensible string of length 4a has a particle A, of mass m, attached at one end and a particle B, of mass m, attached at the other end. The string passes through a small smooth ring which is fixed at a point O at a distance 3a above a horizontal table. The system is hanging in equilibrium with OB=2a when a smooth bead of mass 2m, which is threaded on the string between O and B, is released from rest at O. The bead falls under gravity until it collides with and adheres to the particle B to form a composite particle C.

a) Given that the string remains taut, show that the speed of C immediately after the collision is (ga)^0.5

I did the same question a while back and initially got the same incorrect answer ((4/3)*root(ga)). The answer given in the book, root(ga), is correct.

The mistake we both made is that we forgot to take the particle at the other end, A, into account.

First we calculate the speed of the bead just before it hits B:

v^2 = u^2 + 2as
u = 0, a = g, and s = 2a, so: v^2 = 2g * 2a = 4ga
So: v = 2*root(ga)

Momentum before = momentum of the bead
Momentum after = momentum of the composite particle and the particle A.

I'm using capital V for the new speed:

2mv = (3m + m)V
2mv = 4mV
V = 0.5*v

But: v = 2*root(ga)

So: V = root(ga)

Hope that helps.

Reply 6

baba

~Revol~

Qu.21) A cricket match is played on level ground. Just before the batsman hits the ball, it is 1m above the ground and travelling with speed 20m/s in a direction inclined at 30 degrees below the horizontal. Immediately after the ball has been hit it has speed 25 m/s and the direction of motion is reversed. The ball has mass 0.15kg.

The horizontal distance from the batsman to A (the point where the ball first reaches the ground), is 56.9m.

Find the speed of the ball when it reaches A.

This can be done by using energies, but it can also be done by splitting the velocity into horizontal and vertical components. That may be why so much information is given. I have done it both ways, and I get the same answer. I think energies is a little quicker:

P.E. at start = mgh = 0.15*9.81*1
K.E. at start = 0.5*0.15*(25^2)

When it reaches A there is no P.E., so:

K.E. when it reaches A = 0.5m(v^2) = 0.5*0.15*(v^2) = starting K.E + starting P.E.

So adding both the energies together and dividing by (0.5*0.15) gives v^2.

Square-rooting this gives: 25.39 m/s

Reply 7

baba

~Revol~

Qu 24) This is ONE WEIRD F****** QUESTION! How the **** do you draw the diagram to start with???? :

A smooth horizontal rail is fixed at a height 3h above a rough horizontal ground. A uniform rod AB, of mass M and length 6h, is placed in a vertical plane perpendicular to the rail with the end A resting on the ground. The distance AC = 5h, where C is the point of contact between the rail and the rod. Show that the force exerted by the rail on the rod is of magnitude 12Mg/25.

CAN U ACTUALLY DRAW A DIAGRAM FOR THIS THAT MAKES RELATIVE SENSE WITHOUT?

I've attached my diagram... hope the writing's legible!

m2q24.gif13.6KB

Reply 8

~Revol~

OP

ok, chances are that is the correct diagram!

Who can actually manage to draw that first time, let alone in the exam???? Soz, i am just so annoyed about the difficulty of M2.

Hey Baba, for that energy velocity question, can u show me yer method for doing it with velocities and not energies? coz i attempted it that way, and got a weird answer....

M2 question Review Exercise 2

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