Answer
a) $v_f=0.43m/s$
b) $v_{8kg}=5.0m/s$
$v_{6kg}=-5.5m/s$
c) $v_{8kg}=0.88m/s$
reasonable.
d) $v_{6kg}=1.2m/s$
not reasonable.
e) $v_{8kg}=3.9m/s$
reasonable.
Work Step by Step
a) If the objects stick together, their final velocity must be equal.
$(6.0kg)(6.5m/s)+(8.0kg)(-4.0m/s)=(14kg)v_f$
$v_f=0.43m/s$
b) If collision is elastic:
$(6.0kg)(6.5m/s)+(8.0kg)(-4.0m/s)=(6.0kg)v_a+(8.0kg)v_b$
$\frac{(6.0kg)(6.5m/s)^2}{2}+\frac{(8.0kg)(-4.0m/s)^2}{2}=\frac{(6.0kg)v_a^2}{2}+\frac{(8.0kg)v_b^2}{2}$
$v_b=5.0m/s$ or $v_b=-4.0m/s$
The 8 kg ball is initially moving to the left, so its final velocity must be positive. The correct answer is $v_b=5.0m/s$
$v_a=-5.5m/s$
c) $v_{8kg}=\frac{(6.0kg)(6.5m/s)+(8.0kg)(-4.0m/s)}{8.0kg}=0.88m/s$
d) $v_{6kg}=\frac{(6.0kg)(6.5m/s)+(8.0kg)(-4.0m/s)}{6.0kg}=1.2m/s$
e) $v_{8kg}=\frac{(6.0kg)(6.5m/s)+(8.0kg)(-4.0m/s)-(6.0kg)(-4.0m/s)}{8.0kg}=3.9m/s$
The final velocities were calculated to ensure momentum stays constant, so we should determine final kinetic energy to ensure it is smaller than or equal to initial kinetic energy.
Initial total kinetic energy is: $190J$
Final total kinetic energies are:
c) 3.1J
d) 4.3J
e) 110J
None of the values above are impossible, but d is not possible because if the 8 kg ball is at rest, the 6 kg ball should have a negative final velocity because it was initially moving in the positive direction. c and e are reasonable.
