Question

Block m1 of mass 2m and velocity v0 is traveling to the right (+x) and makes an elastic head-on collision with block m2 of mass m and velocity −2v0 (i.e., traveling to the left). What is the velocity v1′ of block m1 after the collision?

Answer 1

After the elastic collision, block m1 will have a final velocity of -2v0, showing it will move to the left with double its initial speed, while block m2 will have a final velocity of v0.

Step-by-step explanation:

The question asks about the final velocity of block m1 after an elastic collision with block m2. In an elastic collision, both momentum and kinetic energy are conserved. Because block m1 has a mass of 2m and an initial velocity of v0, and block m2 has a mass of m and an initial velocity of -2v0, we can use the conservation of momentum and kinetic energy to find the final velocities.

For conservation of momentum:

m1 * v0 + m2 * (-2v0) = m1 * v1' + m2 * v2'

For conservation of kinetic energy:

(1/2) * m1 * v0^2 + (1/2) * m2 * (-2v0)^2 = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2

Since m1 = 2m and m2 = m, the equations become:

2m * v0 - 2m * v0 = 2m * v1' + m * v2'

m * v0^2 + 2m * v0^2 = m * v1'^2 + m * v2'^2

Solving these equations and assuming an elastic collision, we find that block m1 will have a final velocity of -2v0 and block m2 will end up with a velocity of v0.

Answer 2

1) In any collision the momentum is conserved

(2*m)*(vo) + (m)*(-2*vo) = (2*m)(v1') + (m)(v2')

candel all the m factors (because they appear in all the terms on both sides of the equation)

2(vo) - 2(vo) = 2(v1') + (v2') => 2(v1') + v(2') = 0 => (v2') = - 2(v1')

2) Elastic collision => conservation of energy

=> [1/2] (2*m) (vo)^2 + [1/2](m)*(2*vo)^2 = [1/2](2*m)(v1')^2 + [1/2](m)(v2')^2

cancel all the 1/2 and m factors =>

2(vo)^2 + 4(vo)^2 = 2(v1')^2 + (v2')^2 =>

4(vo)^2 = 2(v1')^2 + (v2')^2

now replace (v2') = -2(v1')

=> 4(vo)^2 = 2(v1')^2 + [-2(v1')]^2 = 2(v1')^2 + 4(v1')^2 = 6(v1')^2 =>

(v1')^2 = [4/6] (vo)^2 =>

(v1')^2 = [2/3] (vo)^2 =>

(v1') = [√(2/3)]*(vo)

Answer: (v1') = [√(2/3)]*(vo)