Time remaining:
physics problem about linear momentum

Physics
Tutor: None Selected Time limit: 1 Day

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?

Apr 13th, 2015


Momentum is always conserved.
Total momentum before = total momentum after
mv = m(v1-v1 after collision) + (3m)(v2-v2 after collision)
Simplify:
v =( v1-v1 after collision)+ 3(v2-v2 after collision)

In elastic collisions, KE is conserved.
Total KE before = total KE after collision
½mv² = ½m(v1-v1 after collision)² + ½(3m)(v2-v2 after collison)²
Simplify:
v² = (v1-v1 after collision)² + 3(v2- v2 after collision)²

therefore:
v =( v1-v1 after collision) + 3(v2_v2 after collision)
v² = (v1-v1 after collision)² + 3(v2_v2after collision)²

Apr 13th, 2015

Momentum is always conserved.
Total momentum before = total momentum after
mv = m(v1-v1 after collision) + (3m)(v2-v2 after collision)
Simplify:
v =( v1-v1 after collision)+ 3(v2-v2 after collision)

In elastic collisions, KE is conserved.
Total KE before = total KE after collision
½mv² = ½m(v1-v1 after collision)² + ½(3m)(v2-v2 after collison)²
Simplify:
v² = (v1-v1 after collision)² + 3(v2- v2 after collision)²

therefore:
v =( v1-v1 after collision) + 3(v2_v2 after collision)

v1=v- 3(v2-v2 after collision) +v1 after collision

v2=v/3-( v1-v1 after collision)/3+v2 after collision


v² = (v1-v1 after collision)² + 3(v2-v2after collision)²

Apr 13th, 2015

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Apr 13th, 2015
...
Apr 13th, 2015
Dec 5th, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer