Help test tmrw elastic collision and conservation of momentum problem?

[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!

 

[Nickie]

In an elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum before the collision is equal to the total momentum after the collision, and the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

In this problem, the mass of block 2 is twice that of block 1. This means that the momentum of block 2 is also twice that of block 1. When the two blocks collide, the momentum from block 1 will be transferred to block 2, resulting in a final momentum of 20 (10 from block 1 and 10 from block 2).

As for why the "m" (mass) term cancels out, this is because both blocks have the same mass (m1 = m2). Therefore, when solving for the final velocities, the mass term can be factored out and canceled, leaving only the velocity terms (v1 and v2).