A 0.400 kg croquet ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball.
(a) What is the mass of the second ball?
(b) What fraction of the original kinetic energy (KE/KE) gets transferred to the second ball?
I know the equation for an elastic collision is (1/2)m1(v1)^2 + (1/2)m2(v2)^2 = (1/2)m1(v1')^2 + (1/2)m2(v2')^2 but I am not exactly sure how to plug the "The second ball moves off with half the original speed of the first ball." part into the equation. And for part B, I know KE is 1/2mv^2 but I have no idea where to begin to answer that question.. if anyone could help that would be great.
Start with the conservation of momentum.
m1V1=m1v1' + m2v2
where v2 is 1/2 V1
Then, solve that equation for V1' in terms of V1, m1, m2 *(you are given m1).
Now, write the conservation of energy..
1/2 m1V1 = 1/2 m1V1' + 1/2 m2V2 (remember v2= 1/2 V1), put for V1' what you got above, then turn the crank on the algebra to solve for m2)
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I don't believe this!
a spring is stretched a distance of 3.6cm when a mass of 8.800kg is hung from it. determine the value of the spring constant in N/m.
a spring is stretched a distance of 3.6cm when a mass of 8.800kg is hung from it. determine the value of the spring constant in N/m.
To solve part (a), we can start with the conservation of momentum equation:
m1V1 = m1v1' + m2v2
Given that the second ball moves off with half the original speed of the first ball, we can substitute v2 = 0.5V1 into the equation:
m1V1 = m1v1' + m2(0.5V1)
Next, we can solve this equation for v1' in terms of V1, m1, and m2:
v1' = (m1V1 - m2(0.5V1))/m1
Now, we can solve for part (b) which asks for the fraction of the original kinetic energy (KE/KE) transferred to the second ball.
The initial kinetic energy is given by:
KE_initial = 0.5m1V1^2
The final kinetic energy is the sum of the kinetic energies of both balls:
KE_final = 0.5m1v1'^2 + 0.5m2v2^2
If we substitute v1' and v2 = 0.5V1 into the equation and simplify, we get:
KE_final = 0.5m1((m1V1 - m2(0.5V1))/m1)^2 + 0.5m2(0.5V1)^2
If we further simplify the equation, we can express KE_final in terms of V1, m1, and m2:
KE_final = 0.75m1V1^2 - 0.5m2V1^2 + 0.0625m2V1^2
Now, we can calculate the fraction of the original kinetic energy transferred to the second ball:
(KE_final/KE_initial) = (0.75m1V1^2 - 0.5m2V1^2 + 0.0625m2V1^2) / (0.5m1V1^2)
Simplifying further, we get:
(KE_final/KE_initial) = (0.75m1 - 0.4375m2) / 0.5m1