I know i did part a correctly. heres what i did: momentum is conserved: m1 * u - m2 * u = m2 * v or (m1 - m2) * u = m2 * v Also, for an elastic head-on collision, we know that the relative velocity of approach = relative velocity of separation (from conservation of energy), or, for this problem, 2u = v Then (m1 - m2) * u = m2 * 2u m1 - m2 = 2 * m2 m1 = 3 * m2 m1 is the sphere that remained at rest (hence its absence from the RHS), so m2 = 0.3kg / 3 m2 = 0.1 kg b) this part confuses me, heres what i did (m1 - m2) * u = m2 * v (.3kg - .1kg)(2.0m/s) = .1kg * v .4 kg = .1 v v = 4 m/s What my teacher did: (.3g - .1g) * 2.0m/s = (.3g + .1g) * v I understand the left hand side but i dont get the right hand side. Why is m1 added to m2 when m1 is at rest which makes its v = zero?? v = +1.00m/s since the answer is positive, what does that mean? Also, if v was -1.00m/s what would that mean? thanks!
<span>Reference https://www.physicsforums.com/threads/elastic-collision-with-conservation-of-momentum-problem.651261...</span>
The answer is refracts parallel to the axis of the lens
potential energy = mass × gravity × height
so, change in potential energy = mass × gravity × change in height
2 = 50 × 10 × Δh
2 ÷ 500 = Δh
Δh = 0.004 m
This distance does depend on the initial velocity of the ball.
Work = force x distance. So the answer would be 200x50=10000 joules
Answer:
F=124.03 N
Explanation:
Given data
mass m=5.00 kg
y(t)=(2.80 m/s)t+(0.61m/s³)t³
To find
Force F at t=4.10 s
Solution
As
derivative with respect to time we get
again derivative with respect to time to get acceleration
at t=4.10 s then a(t) is
a=3.66(4.10)
a=15.006 m/s²
Now to use Newton law to find force in y direction
