- #1
Mangoes
- 96
- 1
Homework Statement
A 10.0 g marble slides to the left with a velocity of magnitude 0.400 m/s on the frictionless, horizontal surface of an icy New York sidewalk and has a head-on, elastic collision with a larger 30.0 g marble sliding to the right with a velocity of magnitude 0.200 m/s.
Find the velocity of each marble after the collision.
The Attempt at a Solution
Since I am ignoring friction, the sum of external forces is zero and momentum is conserved.
[tex] m_1v_{A1} + m_2v_{B1} = m_1v_{A2} + m_2v_{B2} [/tex]
[tex] (0.01 kg)(0.4 m/s) + (0.03 kg)(0.2 m/s) = (0.01 kg)v_A + (0.03 kg)v_B [/tex]
I have two unknowns and one equation.
Since the collision is elastic, kinetic energy is also conserved.
[tex] (1/2)m_1{v_{A1}}^2 + (1/2)m_2{v_{B1}}^2 = (1/2)m_1{v_{A2}}^2 + (1/2)m_2{v_{B2}}^2 [/tex]
[tex] (1/2)(0.1kg)(0.4 m/s)^2 + (1/2)(0.3 kg)(0.2 m/s)^2 = (1/2)(0.1kg){v_{A2}}^2 + (1/2)(0.3kg){v_{B2}}^2 [/tex]
Simplifying the numbers a little and omitting units for simplicity gives the two equations:
[tex] 0.01 = 0.01v_A + 0.03v_B [/tex]
[tex] 0.014 = 0.05{v_A}^2 + 0.15{v_B}^2 [/tex]
I didn't bother going through with the algebra, but plugging in the two equations into WA gives:
http://www.wolframalpha.com/input/?i=0.01+=+0.01x+++0.03y,+0.014+=+0.05x^2+++0.15y^2
Not only do I not have a way to discriminate between which pair of points I should use, but neither of the answers match up. The book tells me that the velocities are vA = 0.5 m/s and vB = -0.1 m/s.
Why isn't this working out?
This exact same thought process worked fine for the problem right before this one.