a question on 3u motion (1 Viewer)

[freaking_out]

a ball is dropped from a lookout 180 metres high. At the same time, a stone is fired vertically upwards from the valley floor with speed V m/s. Take g=10m/s^2

i)Find for what values of V a collision in the air will occur.

ii)Find, in terms of V, the time and the height when collision occurs.

iii)Prove that the collision speed is V m/s.

thanks in advance

 

[wogboy]

i) let Xs = vertical position of the stone,
Xb = vertical position of the ball,
t = time

d^2(Xb)/dt^2 = -10
d(Xb)/dt = -10t (since at t=0, the velocity is 0)
Xb = -5t^2 + 180 (since the lookout is 180m high)

also

d^2(Xs)/dt^2 = -10
d(Xs)/dt = -10t + V (since at t=0, the velocity is V)
Xs = -5t^2 + Vt (when t=0, Xs=0)

First find out how long the ball takes to hit the ground.
0 = -5t^2 + 180
t^2 = 36
t = 6s

Now whenever the collision occurs, Xs = Xb.

-5t^2 + Vt = -5t^2 + 180

V = 180/t

However 0 < t < 6 (the collision "in the air" can't occur AFTER the ball hits the ground)

so V > 30
V > 30 m/s

ii)

V = 180/t (straight from above),
t = 180/V s (this is the time when the collision occurs)

we can put this value of t into the formula for Xb,

Xb = -5t^2 + 180
= -162000/V^2 + 180

(and if you sub t = 180/V into Xs, you should get the same answer)

so x = -162000/V^2 + 180 m (this is the height at which the collision occurs)

iii)

the collision occurs when t=180/V

now d(Xb)/dt = -10t
= -1800/V (when the collision occurs)

and d(Xs)/dt = -10t + V
= -1800/V + V (when the collision occurs)

now v = d(Xs)/dt - d(Xb)/dt,

where v is the relative speed of collision.

so v = -1800/V + V + 1800/V
v = V m/s

so they collide at the speed V m/s.

 

[freaking_out]

hey, can u explain to me this concept of collission speed?

also is this like a typical 3u exam question????

 

[wogboy]

By saying "collision speed", the question really means the relative velocity of collision. The relative velocity of collision of two objects is equal to the "velocity of object1" minus the "velocity of object2", as they collide.

e.g. If an object A moves forward at speed x, and another object B (which is behind A) moves forward with speed y (where y > x), they will collide at speed v given by:

v = y - x

Here's a special case of a collision which is easier to imagine: If an object A moves forward at a certain speed x, and hits a stationary object B. It's obvious from common sense that the speed of collision here is x. This is because the speed of collision is equal to the (velocity of A) - (velocity of B), and since B is still (i.e. has 0 velocity), the speed of collision is simply the speed of A, which is x.

It sounds like a bit more of a physics sort of question, but it could still come up in a 3U exam, so be prepared for this sort of stuff.

 

[freaking_out]

oh yeah, its just the resultant vector(in physics) right???

 

[wogboy]

Yeah sort of. But instead of adding the two vectors together, you need to subtract them (i.e. reverse one of them, and add them), to get their relative velocity. It doesn't matter which vector you reverse (i.e. the order of subtraction), because only the magnitude (absolute value) is required, by the word "speed".

 

[freaking_out]

yep now i get it....thank you- i wonder why they don't show me all these definitions in the text book.

 

[Affinity]

Follow your instincts when things are unfamiliar