Hello, I need this question answerd. no need for work. Thank you. Two 1.5 kg bod
ID: 2256416 • Letter: H
Question
Hello, I need this question answerd. no need for work. Thank you.
Two 1.5 kg bodies, A and B, collide. The velocities before the collision are
and
After the collision,
What are (a) the x-component and (b) the y-component of the final velocity of B? (c) What is the change in the total kinetic energy (including sign)?
thanks again.
Two 1.5 kg bodies, A and B, collide. The velocities before the collision are v rightarrow A = (5.2i + 5.4 j)m/s v rightarrow B = (41i + 1.6j)m/s v rightarrow 'A = (24i + 3.7 j)m/s What are the x-component and the y-component of the final velocity of B? What is the change in the total kinetic energy (including sign)?Explanation / Answer
here momentum is conserved in x and y direction
in x-direction
mA*VAxi + mB*VBxi = mA*VAxf + mB*VBxf
VBxf = (mA*VAxi + mB*VBxi - mA*VAxf)/mB
here mA = mB
VBxf = (VAxi + VBxi - VAxf)
= 5.2 + 41 - 24
= 22.2 m/s <<<<<<----Answer
in x-direction
mA*VAyi + mB*VByi = mA*VAyf + mB*VByf
VByf = (mA*VAyi + mB*VByi - mA*VAyf)/mB
here mA = mB
VByf = (VAyi + VByi - VAyf)
= 5.4 + 1.6 - 3.7
= 3.3 m/s <<<<<<----Answer
Ki = 0.5*mA*(VAxi^2 + VAyi^2) + 0.5*mB*(VBxi^2 + VByi^2)
= 0.5*1.5*(5.2^2 + 5.4^2) + 0.5*1.5*(41^2 + 1.6^2)
= 1304.82
Kf = 0.5*mA*(VAxf^2 + VAyf^2) + 0.5*mB*(VBxf^2 + VByf^2)
= 0.5*1.5*(24^2 + 3.7^2) + 0.5*1.5*(22.2^2 + 3.3^2)
= 820.065
kf - ki = -484.755 J <<<<<<----Answer
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