Answer:
![v_{2f} = \frac{2vm_1}{m_2 + m_1}](https://tex.z-dn.net/?f=v_%7B2f%7D%20%3D%20%5Cfrac%7B2vm_1%7D%7Bm_2%20%2B%20m_1%7D)
Explanation:
If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls
Before the collision
![P_i = m_1v](https://tex.z-dn.net/?f=P_i%20%3D%20m_1v)
After the collision
![P_f = m_1v_{1f} + m_2v_{2f}](https://tex.z-dn.net/?f=P_f%20%3D%20m_1v_%7B1f%7D%20%2B%20m_2v_%7B2f%7D)
So using the law of momentum conservation
![P_i = P_f](https://tex.z-dn.net/?f=P_i%20%3D%20P_f)
![m_1v = m_1v_{1f} + m_2v_{2f}](https://tex.z-dn.net/?f=m_1v%20%3D%20m_1v_%7B1f%7D%20%2B%20m_2v_%7B2f%7D)
We can solve for the speed of ball 1 post collision in terms of others:
![v_{1f} = v - v_{2f}\frac{m_2}{m_1}](https://tex.z-dn.net/?f=v_%7B1f%7D%20%3D%20v%20-%20v_%7B2f%7D%5Cfrac%7Bm_2%7D%7Bm_1%7D)
Their kinetic energy is also conserved before and after collision
![m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2](https://tex.z-dn.net/?f=m_1v%5E2%2F2%20%3D%20m_1v_%7B1f%7D%5E2%2F2%20%2B%20m_2v_%7B2f%7D%5E2%2F2)
![m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2](https://tex.z-dn.net/?f=m_1v%5E2%20%3D%20m_1v_%7B1f%7D%5E2%20%2B%20m_2v_%7B2f%7D%5E2)
From here we can plug in ![v_{1f} = v - v_{2f}\frac{m_2}{m_1}](https://tex.z-dn.net/?f=v_%7B1f%7D%20%3D%20v%20-%20v_%7B2f%7D%5Cfrac%7Bm_2%7D%7Bm_1%7D)
![m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2](https://tex.z-dn.net/?f=m_1v%5E2%20%3D%20m_1%5Cleft%28v%20-%20v_%7B2f%7D%5Cfrac%7Bm_2%7D%7Bm_1%7D%5Cright%29%5E2%20%2B%20m_2v_%7B2f%7D%5E2)
![m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2](https://tex.z-dn.net/?f=m_1v%5E2%20%3D%20m_1%5Cleft%28v%5E2%20-%202vv_%7B2f%7D%5Cfrac%7Bm_2%7D%7Bm_1%7D%20%2B%20v_%7B2f%7D%5E2%5Cfrac%7Bm_2%5E2%7D%7Bm_1%5E2%7D%5Cright%29%20%2B%20m_2v_%7B2f%7D%5E2)
![m_1v^2 = m_1v^2 - 2vv_{2f}m_2 + v_{2f}^2\frac{m_2^2}{m_1} + m_2v_{2f}^2](https://tex.z-dn.net/?f=m_1v%5E2%20%3D%20m_1v%5E2%20-%202vv_%7B2f%7Dm_2%20%2B%20v_%7B2f%7D%5E2%5Cfrac%7Bm_2%5E2%7D%7Bm_1%7D%20%2B%20m_2v_%7B2f%7D%5E2)
![v_{2f}^2(m_2 + \frac{m_2^2}{m_1}) - 2vm_2v_{2f} = 0](https://tex.z-dn.net/?f=v_%7B2f%7D%5E2%28m_2%20%2B%20%5Cfrac%7Bm_2%5E2%7D%7Bm_1%7D%29%20-%202vm_2v_%7B2f%7D%20%3D%200)
![v_{2f}(1 + \frac{m_2}{m_1}) = 2v](https://tex.z-dn.net/?f=v_%7B2f%7D%281%20%2B%20%5Cfrac%7Bm_2%7D%7Bm_1%7D%29%20%3D%202v)
![v_{2f} = \frac{2v}{1 + \frac{m_2}{m_1}} = \frac{2v}{\frac{m_1 + m_2}{m_1}} = \frac{2vm_1}{m_2 + m_1}](https://tex.z-dn.net/?f=v_%7B2f%7D%20%3D%20%5Cfrac%7B2v%7D%7B1%20%2B%20%5Cfrac%7Bm_2%7D%7Bm_1%7D%7D%20%3D%20%5Cfrac%7B2v%7D%7B%5Cfrac%7Bm_1%20%2B%20m_2%7D%7Bm_1%7D%7D%20%3D%20%5Cfrac%7B2vm_1%7D%7Bm_2%20%2B%20m_1%7D)
Speed is the distance traveled
time
but velocity is the change in distance
time
Answer:
Final velocity = 7.677 m/s
KE before crash = 202300 J
KE after crash = 182,702.62 J
Explanation:
We are given;
m1 = 1400 kg
m2 = 4700 kg
u1 = 17 m/s
u2 = 0 m/s
Using formula for inelastic collision, we have;
m1•u1 + m2•u2 = (m1 + m2)v
Where v is final velocity after collision.
Plugging in the relevant values;
(1400 × 17) + (4700 × 0) = (1400 + 1700)v
23800 = 3100v
v = 23800/3100
v = 7.677 m/s
Kinetic energy before crash = ½ × 1400 × 17² = 202300 J
Kinetic energy after crash = ½(1400 + 1700) × 7.677² = 182,702.62 J
For the first part of this question, consider that "weight" can be described as mass x acceleration of gravity. Weight is expressed in Newtons. To solve for mass in this case, simply divide 9800N by 9.8m/s^2 (Earth's gravitational acceleration). This will give you a mass of 1000 kg. This mass is moved due to the net force supplied by the normal force from the rocket "pushing" off of Earth.
For the second part, we will use the equation F = ma, which is Newton's second law. For this, we know the m, or mass, is 1000 kg. Also, we know the a, or acceleration, will be 4 m/s^2. To solve for force, we will multiply both of these values. This gives a force of 4000 N. I hope this clears things up!
Answer:
![Q = \frac{0.068}{E}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B0.068%7D%7BE%7D)
where E = electric field intensity
Explanation:
As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical
So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field
while weight of the ball is vertically downwards
so here we have
![QE = F_x](https://tex.z-dn.net/?f=QE%20%3D%20F_x)
![mg = F_y](https://tex.z-dn.net/?f=mg%20%3D%20F_y)
since string makes 30 degree angle with the vertical so we will have
![tan\theta = \frac{F_x}{F_y}](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%20%5Cfrac%7BF_x%7D%7BF_y%7D)
![tan30 = \frac{QE}{mg}](https://tex.z-dn.net/?f=tan30%20%3D%20%5Cfrac%7BQE%7D%7Bmg%7D)
![Q = \frac{mg}{E}tan30](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7Bmg%7D%7BE%7Dtan30)
![Q = \frac{0.012\times 9.81}{E} tan30](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B0.012%5Ctimes%209.81%7D%7BE%7D%20tan30)
![Q = \frac{0.068}{E}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B0.068%7D%7BE%7D)
where E = electric field intensity