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Mechanics questions
By Peter Warner (P3272) on Saturday, November 18, 2000 - 11:35 am:

Hi,

I am stuck on part (b) of this question. And I am not sure if I have answered part (a) completely.
It would be helpful if someone could help out.

At time t=0, two ice skaters John (J) and Norma (N) have position vectors 40j metres and 20i metres relative to an origin 0 at the centre of an ice rink, where i and j are unit vectors perpendicular to each other. John has constant velocity (3i + 4j)m/sec.

a) show that the skaters will collide and find the time at which the collision takes place

b) on another occasion, John has position vector 40j metres. he wises to skate in a straight line to the point with position vector 30i metres. given that his speed is constant at 5 m/sec, find his velocity.

a)
(J) x = (5t)i + (40)j
(N) x = (3t + 20)i + (4t)j

equate i coefficients
(5t) = (3t + 20)
t = 10, they collide at 10 seconds,

* does this answer part (a) completely?

Thanks
Pete
By Hal 2001 (P3046) on Saturday, November 18, 2000 - 11:43 am:

Pete,

Velocity = disp/time = (30i - 40j)/time
But Time = Dist/Speed = sqr(40×40 + 30×30) / 5 = 50/5 = 10 sec
Hence, Vel = 30i - 40j/10 = 3i - 4j

By Peter Warner (P3272) on Saturday, November 18, 2000 - 02:19 pm:

I am stuck on the question below, can anyone help?

A particle P moves in a straight line with constant retardation. At the instants when P passed through the points A, B, C it was moving with speeds 10 m/sec, 7 m/sec and 3 m/sec respectively. Prove that (AB/BC)=(51/40).

Thanks
By Kerwin Hui (Kwkh2) on Saturday, November 18, 2000 - 07:25 pm:

Using the equation

v2=u2+2as

over the interval AB and BC, we have

-2a(AB)=102-72
-2a(BC)=72-32

Divide the two equations and the answer comes out nicely.

Kerwin

By Peter Warner (P3272) on Saturday, November 18, 2000 - 09:01 pm:

Neat! Thanks Kerwin!

By Peter Warner (P3272) on Sunday, November 19, 2000 - 05:20 pm:

Hi,

I tried to tackle the following question, but kept hitting hurdles. Can someone help out here? Thanks.

A ball is projected from a point A on horizontal ground, with speed 14 m/sec at an angle a degrees, and moves freely under gravity. At an instant when the ball is at a point P, which is 5m above the ground , the components of velocity of the ball horizontally and vertically upwards are both u m/sec.

a) By using energy considerations, or otherwise, show that u=7.

b) Obtain the value of a

c) Find in seconds to 2 dp the time taken by the ball to reach its greatest height above the ground.

d) The diagram shows a fixed plane inclined at an angle of 50 degrees to the horizontal. The ball strikes this plane, at right angles with speed V m/sec. Calculate the value of V, to 1 dp.

a)
PE gained at 5 metres height, PE = mgh = 5mg Joules
KE lost = 0.5mu2 - 0.5mv2

Got a bit lost after this..

Thanks for any help.
Pete
By Kerwin Hui (Kwkh2) on Monday, November 20, 2000 - 12:21 pm:

The velocity of the ball at A is
14cosa m/s horizontally
14sina m/s vertically upwards

Apply v2=u2+2as to the vertical component, and remember that the horizontal component is unchanged, part (a) and (b) follow quite nicely.

Part (c) should be OK when you know the value of a and hence the initial vertical velocity.

Part (d) requires you to find out the vertical component of velocity and thus the speed.

Kerwin

By Peter Warner (P3272) on Monday, November 20, 2000 - 02:47 pm:

Hal, thanks for the earlier help.

Kerwin, thank you for that help also.
I shall now try to tackle the question using the advice you have given.

Pete