A 0.500 kg sphere moving with a velocity (2.00 - 3.40 + 1.00) m/s strikes anothe

Question

A 0.500 kg sphere moving with a velocity (2.00 - 3.40 + 1.00) m/s strikes another sphere of mass 1.50 kg moving with a velocity (-1.00 + 2.00 - 2.70) m/s. a) If the velocity of the 0.500 kg sphere after the collision is (-0.80i + 3.00j - 8.00k) m/s, find the velocity of the 1.50 kg sphere and identify the kind of collision (elastic, inelastic, or perfectly inelastic). (b) If the velocity of the 0.500 kg sphere after the collision is (-0.250 + 0.650 - 1.78) m/s, find the final velocity of the 1.50 kg sphere and identify the kind of collision. (c) If the velocity of the 0.500 kg sphere after the collision is (-1.00i + 2.60j + ak) m/s, find the value of a and the velocity of the 1.50 kg sphere after an elastic collision. (Two values of a are possible, a negative value and a positive value. Report each with their corresponding final velocities.) a (positive value) ________ m/s2 v2f = ____________ m/s a (negative value) ________ m/s2 v2f = ____________ m/s I know how to do a and b. i just need help with c.

Explanation / Answer

part c) conservation of momentum 0.5* (2 i + 3.4 j + 1 k) + 1.5 * (-1 i + 2 j -2.7 k) = 0.5*(-1i +2.6j - a k ) + 1.5( x i + y j + z k) so this means 1 - 1.5 = -0.5 + 1.5*x x = 0 also means 0.5*3.4 + 1.5*2 = 0.5*2.6 + 1.5y y=2.27 m/s and in z direction 0.5 + 1.5*-2.7 = -a/2 + 1.5 z so 1 + 3*-2.7 = -a + 3z a = 3z - 1 +3*2.7 now use conservation of energy 0.5*(2^2 + 3.4^2 + 1^2) + 1.5*(1^2+2^2 + 2.7^2) = 0.5*(1^2+2.6^2 + a^2 )+1.5( x^2 + y^2 + z^2) 0.5*(2^2 + 3.4^2 + 1^2) + 1.5*(1^2+2^2 + 2.7^2) = 0.5*(1^2+2.6^2 + (3z - 1 +3*2.7)^2 )+1.5( 0^2 + 2.27^2 + z^2) z=-2.99 or -0.564 m/s so a= -1.87 or 5.408 m/s

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