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>
Answer:
potential energy PE = M g h
KE at bottom = 1/2 M V^2
Regardless of the slope of the slide the change in energy is the same
1/2 V^2 = g h
V = (2 g h)^1/2 = (2 * 9.8 m/s^2 * 10 m)^1/2 = 14 m / s
Perhaps the question says that h = 55 * .1 = 5.5 m
Then V = (2 * 9.8 * 5.5) = 10.4 m/s
Anions: negatively charged atoms
Cations: positively charged atoms (CAT-ions are PAWS-itively charged)
Anode: positively charged electrode that allows electrical current flow (positive end of battery)
Cathode: negatively charged electrode that allows electrical current flow (negative end of batter)
Answer:
3984875 N/C
Explanation:
Applying,
E = kq/r².................. Equation 1
Where E = Electric field, q = charge, r = distance, k = coulomb's constant.
From the question,
Given: q = 7.10 mC = 0.0071 C, r = 4.0 m
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values into equation 1
E = 0.0071(8.98×10⁹)/4²
E = 3984875 N/C
Hence the electric field is 3984875 N/C
1. sliding 2. static 4. rolling