physics problem about linear momentum
Physics

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particle 1 of mass M movies with speed v in the +x direction and has an elastic collision with particle 2 (mass=3M) that was orginally at rest. after the collision, particle 2 is moving in the +x direction. what is its speed?
Momentum is always conserved.
Total momentum before = total momentum after
mv = m(v1v1 after collision) + (3m)(v2v2 after collision)
Simplify:
v =( v1v1 after collision)+ 3(v2v2 after collision)
In elastic collisions, KE is conserved.
Total KE before = total KE after collision
½mv² = ½m(v1v1 after collision)² + ½(3m)(v2v2 after collison)²
Simplify:
v² = (v1v1 after collision)² + 3(v2 v2 after collision)²
therefore:
v =( v1v1 after collision) + 3(v2_v2 after collision)
v² = (v1v1 after collision)² + 3(v2_v2after collision)²
Momentum is always conserved.
Total momentum before = total momentum after
mv = m(v1v1 after collision) + (3m)(v2v2 after collision)
Simplify:
v =( v1v1 after collision)+ 3(v2v2 after collision)
In elastic collisions, KE is conserved.
Total KE before = total KE after collision
½mv² = ½m(v1v1 after collision)² + ½(3m)(v2v2 after collison)²
Simplify:
v² = (v1v1 after collision)² + 3(v2 v2 after collision)²
therefore:
v =( v1v1 after collision) + 3(v2_v2 after collision)
v1=v 3(v2v2 after collision) +v1 after collision
v2=v/3( v1v1 after collision)/3+v2 after collision
v² = (v1v1 after collision)² + 3(v2v2after collision)²
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