Elastic collision: why does mass cancel in momentum conservation?

[nchin]

Block 1 moves with speed of 10m/s to right. It hits block 2 which has twice the mass of block 1 and speed of 5m/s to right. compute the magnitude and direction of block 1 for a perfectly elastic collision.

solution:

u1 = 10m/s
u2 = 5m/s

m1v1 + m2v2 = m1u1 + m2u2 ---->
v1 + 2v2 = u1 + 2u2
... = 10 + 2(5) = 20

(1) v1 + 2v2 = 20

why did u2 and v2 get multiplied by two and why did the m cross out?

 

[doub]

I not certain but I think I would go about solving it like this;

The formula you want to use is v'_A = ((m_A-m_B)/(m_A + m_B))v_A

And look at the frame set so that the total velocity of block 1 is actual 10 m/s - 5 m/s so the sped of block one would be 5 m/s.

No guarantee though

 

[ehild]

Block 1 moves with speed of 10m/s to right. It hits block 2 which has twice the mass of block 1 and speed of 5m/s to right. compute the magnitude and direction of block 1 for a perfectly elastic collision.

solution:

u1 = 10m/s
u2 = 5m/s

m1v1 + m2v2 = m1u1 + m2u2 ---->
v1 + 2v2 = u1 + 2u2
... = 10 + 2(5) = 20

(1) v1 + 2v2 = 20

why did u2 and v2 get multiplied by two and why did the m cross out?

The second block has twice as much as the first one. So you can write for the mass of the first block "m" and "2m" for the mass of the second one. As all terms in the equation have the common factor m you can cross it out.

ehild

 

[nchin]

The second block has twice as much as the first one. So you can write for the mass of the first block "m" and "2m" for the mass of the second one. As all terms in the equation have the common factor m you can cross it out.

ehild

thanks!