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November 24, 2014

November 24, 2014

Posted by **Shaila** on Sunday, September 26, 2010 at 2:52pm.

Ans: since it is 2D collision, i tired solving it before and after; vertical component and horizontal component, i keep getting my second angle as 54deg and velocity 7m/s, where as my answer should be 41deg and 6m/s.

Before

Vertical

0.155(12.5)

=1.9375

Horizontal = 0

After

Vertical

(0.155)(9.56)(cos 29.7) + (0.155)(V'sin Theta) = 1.9375

(0.155)(V'sin Theta) = 1.9375 - (0.155)(9.56)(cos 29.7) (eq 1)

Horizontal

(0.155)(9.56)(V' cos 29.7) -(0.155)(V' cos theta) = 0

(0.155)(V' cos theta) = -(0.155)(9.56)(V' cos 29.7) (eq 2)

Then you solve simultaneously, but i do not get the answer

- Physics - Inelastic Collision ( check + help) -
**MathMate**, Sunday, September 26, 2010 at 3:24pmWe do not know if it is an elastic collision, but momentum is always conserved.

Your calculations seem to be based on the conservation of momentum in both directions, but you had a cos(θ) for both the "horizontal", or x-direction, and the "vertical", or y-direction.

Assuming the original direction of the first ball is along the y-axis, then the deviation of its course, θ, is -29.7°, and the new velocity, v1=9.56 m/s.

The stationary ball now moves with a velocity v2, and at an angle φ with the y-axis.

Equate momentum in the x-direction:

v1*sin(θ)+v2*sin(φ)=0

Equate momentum in the y-direction:

v1*cos(θ)+v2*cos(φ)=12.5

Solving the equations, I get v2=6.3 and φ=48°, different from your answer.

After that, you equate initial KE and final KE to determine if the collision was elastic.

- Physics - Inelastic Collision ( check + help) -
**Shaila**, Sunday, September 26, 2010 at 8:01pmHow is it possible that mass which is 0.155kg is not taken into consideration?

You solve it as a 2D collision and thereafter you obtain the two equations, two variables and you solve you will get V'

Thereafter you find if it is elastic or not by equating the kinetic energy

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