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:
The angular momentum of the particle is 58.14 kg m²/s along positive z- axis and is independent of time .
Explanation:
Given that,
Mass = 1.70 kg
Position vector ![r=(6.00\hat{i}+5.70 t \hat{j})](https://tex.z-dn.net/?f=r%3D%286.00%5Chat%7Bi%7D%2B5.70%20t%20%5Chat%7Bj%7D%29)
We need to calculate the angular velocity
The velocity is the rate of change of the position of the particle.
![v = \dfrac{dr}{dt}](https://tex.z-dn.net/?f=v%20%3D%20%5Cdfrac%7Bdr%7D%7Bdt%7D)
![v=\dfrac{d}{dt}(6.00\hat{i}+5.70 t \hat{j})](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bd%7D%7Bdt%7D%286.00%5Chat%7Bi%7D%2B5.70%20t%20%5Chat%7Bj%7D%29)
![v=5.70\hat{j}](https://tex.z-dn.net/?f=v%3D5.70%5Chat%7Bj%7D)
We need to calculate the angular momentum of the particle
Using formula of angular momentum
![L=r\cdot p](https://tex.z-dn.net/?f=L%3Dr%5Ccdot%20p)
Where, p = mv
Put the value of p into the formula
![L=m(r\times v)](https://tex.z-dn.net/?f=L%3Dm%28r%5Ctimes%20v%29)
Substitute the value into the formula
![L=1.70(6.00\hat{i}+5.70 t \hat{j}\times5.70\hat{j})](https://tex.z-dn.net/?f=L%3D1.70%286.00%5Chat%7Bi%7D%2B5.70%20t%20%5Chat%7Bj%7D%5Ctimes5.70%5Chat%7Bj%7D%29)
![L=1.70\times34.2](https://tex.z-dn.net/?f=L%3D1.70%5Ctimes34.2)
![L=58.14\ kgm^2/s](https://tex.z-dn.net/?f=L%3D58.14%5C%20kgm%5E2%2Fs)
Hence, The angular momentum of the particle is 58.14 kg m²/s along positive z- axis and is independent of time .
Answer:
5000 W
Explanation:
you are technically correct, but the question is asking for units in Watts
Recall that Power rating is the energy per unit time
= Energy ÷ Time
= 300,000 ÷ 60
= 5000 j/s
but we are told that the unit is a single letter, and recall that joules/s = watts
hence
5000j/s = 5000 W
edit: typo. Missing a zero from answer
Answer:
6.5 m above the floor and 5 m above Christine's hand when it reaches the maximum height.
Explanation:
Let g = 10 m/s2 be the gravitational deceleration that affects the ball vertical motion so it comes to the maximum height at 0 speed. We can use the following equation of motion to find out the distance traveled by the ball from where it's thrown:
![v^2 - v_0^2 = 2g\Delta s](https://tex.z-dn.net/?f=v%5E2%20-%20v_0%5E2%20%3D%202g%5CDelta%20s)
where v = 0 m/s is the final velocity of the ball when it reaches maximum level,
= 10m/s is the initial velocity of the ball when it starts, g = -10 m/s2 is the deceleration, and
is the distance traveled, which we care looking for:
![0^2 - 10^2 = -2*(-10)\Delta s](https://tex.z-dn.net/?f=0%5E2%20-%2010%5E2%20%3D%20-2%2A%28-10%29%5CDelta%20s)
![\Delta s = 100 / (2 * 10) = 5 m](https://tex.z-dn.net/?f=%5CDelta%20s%20%3D%20100%20%2F%20%282%20%2A%2010%29%20%3D%205%20m)
So the ball is 5 m above Christine' hands when it reaches maximum height, and since the hand is 1.5 m above the floor, the ball is 5 + 1.5 = 6.5 m above the floor when it reaches maximum height.