Projectile Motion MIF Q25 (1 Viewer)

jkerr138 · Aug 14, 2015

Having a problem with Q25 part b)
I'll provide the whole question below.

'A ball is thrown up with a velocity of 15ms^-1 at an angle of 60 degrees. At the same time, another ball is thrown towards the first one, at an angle of 45 degrees and a velocity of 20ms^-1. The two balls are 30m apart.

a) How far apart will they land? - (Ans 29.9m) Fine with this section
b) Where will the two balls cross over? Will they collide? (Use g= 10ms^-2).

with part b) I thought of setting the two X equations equal to each other and placing 30-10sqrt2t on the RHS, but I'm not sure if it works, so then I made the assumption that they do collide by setting the Y equations equal to find the time, but it sounds a bit off track.

Soln part b) X= 11.5 y=8.1 from memory for Y.

Thanks.

InteGrand · Aug 14, 2015

jkerr138 said:
Having a problem with Q25 part b)
I'll provide the whole question below.

'A ball is thrown up with a velocity of 15ms^-1 at an angle of 60 degrees. At the same time, another ball is thrown towards the first one, at an angle of 45 degrees and a velocity of 20ms^-1. The two balls are 30m apart.

a) How far apart will they land? - (Ans 29.9m) Fine with this section
b) Where will the two balls cross over? Will they collide? (Use g= 10ms^-2).

with part b) I thought of setting the two X equations equal to each other and placing 30-10sqrt2t on the RHS, but I'm not sure if it works, so then I made the assumption that they do collide by setting the Y equations equal to find the time, but it sounds a bit off track.

Soln part b) X= 11.5 y=8.1 from memory for Y.

Thanks.

To find where the balls cross over, you can equate the x(t) equations for each ball and solve for t, then sub. this back in to either x equation to get the value of x(t) at the crossover point.

You can sub. this t value into the y(t) equations for each ball. If they have the same y value at this time, they collide, otherwise, they don't collide.

jkerr138 · Aug 14, 2015

I did try to do this before I placed it on the forum, but you end up with 7.5t-10sqrt2 t= 0
I thought of reusing the 30m and forming the expression: 7.5t=30-10sqrt2 , but I wasn't sure if I am allowed to reuse it?

InteGrand · Aug 14, 2015

jkerr138 said:
I did try to do this before I placed it on the forum, but you end up with 7.5t-10sqrt2 t= 0
I thought of reusing the 30m and forming the expression: 7.5t=30-10sqrt2 , but I wasn't sure if I am allowed to reuse it?

$The equation $7.5t-10\sqrt{2}t=0$ shouldn't be right, since this would imply they also have the same $x$-value at $t=0$, which is not the case, since they start off at different $x$-values. (I haven't attempted the Q.)$

jkerr138 · Aug 15, 2015

InteGrand said:
$The equation $7.5t-10\sqrt{2}t=0$ shouldn't be right, since this would imply they also have the same $x$-value at $t=0$, which is not the case, since they start off at different $x$-values. (I haven't attempted the Q.)$

I would appreciate it if you had an opportunity to solve the problem, as you previously mentioned setting the X(t) equations equal to each other, which I had done before-hand. You disputed this in my previous post with my calculations. Could you extend on your clarifications please? Thanks.

enigma_1 · Aug 15, 2015

InteGrand · Aug 15, 2015

jkerr138 said:
I would appreciate it if you had an opportunity to solve the problem, as you previously mentioned setting the X(t) equations equal to each other, which I had done before-hand. You disputed this in my previous post with my calculations. Could you extend on your clarifications please? Thanks.

$I think you made a mistake with the equations of motion, which should be done with respect to a fixed origin. So we shall define the origin. Let the ball that starts with $15 \text{ m s}^{-1}$ velocity start at the origin $O$ (in other words, define the origin to be this ball starts). Let the other ball start on its right, so it starts at $x=30$. Let the ball that starts at the origin be $B_1$ and the other ball be $B_2$. Using subscript 1's to denote equations for $B_1$, and similarly for $B_2$, you should obtain these equations for the $x$-motion as a function of time $t\geq 0$:$

$x_1 (t) = 15 \cos 60^\circ t \Longleftrightarrow x_1 (t) = \frac{15}{2}t$

$$x_2 (t) = 30 - 20\cos 45^\circ t \Longleftrightarrow x_2 (t) = 30 - 10\sqrt{2}t$.$

$You should now have no problems with equating the equations of motion.$

jkerr138 · Aug 16, 2015

InteGrand said:
$I think you made a mistake with the equations of motion, which should be done with respect to a fixed origin. So we shall define the origin. Let the ball that starts with $15 \text{ m s}^{-1}$ velocity start at the origin $O$ (in other words, define the origin to be this ball starts). Let the other ball start on its right, so it starts at $x=30$. Let the ball that starts at the origin be $B_1$ and the other ball be $B_2$. Using subscript 1's to denote equations for $B_1$, and similarly for $B_2$, you should obtain these equations for the $x$-motion as a function of time $t\geq 0$:$

$x_1 (t) = 15 \cos 60^\circ t \Longleftrightarrow x_1 (t) = \frac{15}{2}t$

$$x_2 (t) = 30 - 20\cos 45^\circ t \Longleftrightarrow x_2 (t) = 30 - 10\sqrt{2}t$.$

$You should now have no problems with equating the equations of motion.$

Thanks for your help, I've figured it out. Poorly worded second part of the question. It had to be turned into cartesian equations ie) Eliminate parameter t in both x(t) equations and substitute into each respective Y equations to solve for X, then sub into y. I guess you could say you were sort of on the way.

jkerr138 · Aug 16, 2015

InteGrand said:
$I think you made a mistake with the equations of motion, which should be done with respect to a fixed origin. So we shall define the origin. Let the ball that starts with $15 \text{ m s}^{-1}$ velocity start at the origin $O$ (in other words, define the origin to be this ball starts). Let the other ball start on its right, so it starts at $x=30$. Let the ball that starts at the origin be $B_1$ and the other ball be $B_2$. Using subscript 1's to denote equations for $B_1$, and similarly for $B_2$, you should obtain these equations for the $x$-motion as a function of time $t\geq 0$:$

$x_1 (t) = 15 \cos 60^\circ t \Longleftrightarrow x_1 (t) = \frac{15}{2}t$

$$x_2 (t) = 30 - 20\cos 45^\circ t \Longleftrightarrow x_2 (t) = 30 - 10\sqrt{2}t$.$

$You should now have no problems with equating the equations of motion.$

Also didn't require the 30m. Just X=10sqrt 2 and x=7.5t

InteGrand · Aug 16, 2015

jkerr138 said:
Also didn't require the 30m. Just X=10sqrt 2 and x=7.5t

$What do you mean didn't require it? In my method, it is incorporated into the $x_2(t)$ equation. You need to use the 30 m in some way, since it clearly affects the answer (imagine if the distance were 1000 m or something instead, the distance they are apart at the start has an effect).$

jkerr138 · Aug 16, 2015

InteGrand said:
$What do you mean didn't require it? In my method, it is incorporated into the $x_2(t)$ equation. You need to use the 30 m in some way, since it clearly affects the answer (imagine if the distance were 1000 m or something instead, the distance they are apart at the start has an effect).$

Only in part a) 30m is required to determine the distance between the two objects.
Balls are 19.5+40-30m= 29.5m (soln)

If your not completely satisfied, have a crack at it and see if you can reach the same answer X= 11.5, y= 8.2 for B)
I'd be interested to see if it does.

InteGrand · Aug 16, 2015

jkerr138 said:
Only in part a) 30m is required to determine the distance between the two objects.
Balls are 19.5+40-30m= 29.5m (soln)

If your not completely satisfied, have a crack at it and see if you can reach the same answer X= 11.5, y= 8.2 for B)
I'd be interested to see if it does.

How did you do it without using the 30 m? You must have used it somehow, because if for example the balls were 10000 m away to start with, they wouldn't even cross each other in their flight path.

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