Asked by brett

A 2.00 kg ball moving to the right at 10.0 m/s makes an off-center collision with a stationary 3.00 kg ball. After the collision, the 2.00 kg ball is deflected upward at an angle of 30o from its original direction of motion and the 3.0 kg ball is moving at 4.0 m/s. Find the speed of the 2.0 kg ball and the direction of the 3.0 kg ball after the collision.

so far i have come up with
Px:(m1v1'costheta)+(m2V2'cosphi) = m1V1
PY:(m1vi'sintheta)-(m2v2'sinphi)=0

cosphi=(m1v1-m1v1'costheta)/m2v2'
sinphi=m1v1'sintheta/m2v2'

Answers

Answered by Henry
M1*V1 + M2*V2 = M1*V3 + M2*V4,
2*10 + 3*0 = 2*V3[30o] + 3*4,
20 = 2V3[30o] + 12,
2V3[30] = 8.
2[30] = 8/V3, 1.732 + i = 8/V3,
8/V3 = 2, 2V3 = 8, V3 = 4 m/s. = Speed of 2kg ball.



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