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Momentum question, been stuck for days now!!

I've been looking through the text book and some tricky mechanics questions for M1 and I am frustrated by my lack of ability to do this one, if someone could be kind enough to explain it to me I'd be more than grateful.

Two smooth spheres A and B, of equal radii and masses m1 and m2 respectively, are moving towards each other along the same horizontal line each with speed u. After collision both spheres have reversed their original directions of motion and A now travels with speed u/2.
Show that 3m1>2m2
Sphere B then strikes a fixed wall at right angles, the impact reversing the direction of motion of B and halving it's speed. Show that B will again collide with A provided that 3m1>4m2.
Reply 1
Original post by Lucykathryn
I've been looking through the text book and some tricky mechanics questions for M1 and I am frustrated by my lack of ability to do this one, if someone could be kind enough to explain it to me I'd be more than grateful.

Two smooth spheres A and B, of equal radii and masses m1 and m2 respectively, are moving towards each other along the same horizontal line each with speed u. After collision both spheres have reversed their original directions of motion and A now travels with speed u/2.
Show that 3m1>2m2


Conservation of linear momentum: um1um2=um12+m2vbum_1 - um_2 = -\frac{um_1}{2} + m_2 v_b

So: um1(1+12)=m2(u+vb)    um1=2m23(u+vb)    3m1=2m2u(u+vb)um_1 \left(1 + \frac{1}{2}\right) = m_2 (u + v_b) \iff um_1 = \frac{2m_2}{3}(u+v_b) \iff 3m_1 = \frac{2m_2}{u}(u+v_b)

So...?

Remember that u+vbu=1+something>1\frac{u+v_b}{u} = 1 + \text{something} > 1.

So 3m1=2m2×(something bigger than 1)3m_1 = 2m_2 \times (\text{something bigger than 1})
(edited 8 years ago)
Reply 2
Original post by Zacken
Conservation of linear momentum: um1um2=um12+m2vbum_1 - um_2 = -\frac{um_1}{2} + m_2 v_b

So: um1(1+12)=m2(u+vb)    um1=2m23(u+vb)    3m1=2m2u(u+vb)um_1 \left(1 + \frac{1}{2}\right) = m_2 (u + v_b) \iff um_1 = \frac{2m_2}{3}(u+v_b) \iff 3m_1 = \frac{2m_2}{u}(u+v_b)

So...?


I am really unsure of how to decide whether a collision will occur just from the momentum. I managed to prove the first part.
Sorry I should have specified where I needed help
Reply 3
Original post by Lucykathryn
I am really unsure of how to decide whether a collision will occur just from the momentum. I managed to prove the first part.
Sorry I should have specified where I needed help


Sphere B will collide with Sphere A if and only if the speed of Sphere B after colliding with the wall is greater than the Speed of A, this then means that B will be able to "catch up" with A. So find an expression for the velocity of B and set it greater than u/2.
Reply 4
Original post by Zacken
Sphere B will collide with Sphere A if and only if the speed of Sphere B after colliding with the wall is greater than the Speed of A, this then means that B will be able to "catch up" with A. So find an expression for the velocity of B and set it greater than u/2.


Oh god that is so simple. Thankyou so much, I over complicated that loads.
Reply 5
Original post by Lucykathryn
Oh god that is so simple. Thankyou so much, I over complicated that loads.


No problem, let me know if it works out! :smile:
Reply 6
Original post by Zacken
No problem, let me know if it works out! :smile:

It worked perfectly, thankyou :smile:
Reply 7
Original post by Lucykathryn
It worked perfectly, thankyou :smile:


Awesome, glad I could help! :h:

Welcome to TSR, hope to see you around the maths forum more. :smile:

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