When analyzing inelastic collisions, we need to consider the law of conservation of momentum, which states that the total momentum, p, of the closed system is a constant. In the case of inelastic collisions, the momentum of the combined mass after the collision is equal to the sum of the momentum of each of the initial masses.
p1+p2+...=pf
In our case we only have two masses, which makes our problem fairly simple. Lets plug in the formula for momentum; p=mv.
m1v1+m2v2=(m1+m2)vf
To find the velocity of the combined mass we simply rearrange the terms.
vf=m1v1+m2v2m1+m2
Plug in the values given in the problem.
vf=(3.0kg)(1.4m/s)+(2.0kg)(0m/s)03.0kg+2.0kg
vf=.84m/s
Answer:
While controversial at first, Thomson's discoveries were gradually accepted by scientists. Eventually, his cathode ray particles were given a more familiar name: electrons. The discovery of <em>the electron</em> disproved the part of Dalton's atomic theory that assumed atoms were indivisible.
Answer:
I HOPE THIS IS CORRECT
Explanation:
Power of water =2 kw=2000w
Mass of water =200kg
difference in temperature ΔT=70−10=60oC
Concept
energy required to heat the water = energy given by water in time t=pt
energy required to increase tempeature of water by 60oC,Q=msΔT
S= specific heat =4200J/kgoC
pt=msΔT
2000×t=200×4200×60
t=25200
or t=25.2×103sec.
Answer:
(a)
vf_1s = 5.19 m/s
h_1s = 10.095 m
vf_4s = 24.23 m/s
h_4s = 4.49 m (below railing)
(b)
vi = 9.9 m/s
(c)
t = 1.53 s
h = 34.41 m
Explanation:
(c)
First, we will use the 1st equation of motion to find the time to attain the highest point:
![v_{f} = v_{i} + gt\\t = \frac{v_{i} - v_{f}}{g}\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3D%20v_%7Bi%7D%20%2B%20gt%5C%5Ct%20%3D%20%5Cfrac%7Bv_%7Bi%7D%20-%20v_%7Bf%7D%7D%7Bg%7D%5C%5C)
where,
t = time to attain maximum height = ?
vf = final velocity = 0 m/s (ball momentarily stops at highest point
vi = initial velocity = 15 m/s
g = - 9.81 m/s (for upward motion)
![t = \frac{0\ m/s - 15\ m/s}{- 9.81\ m/s^2}\\](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B0%5C%20m%2Fs%20-%2015%5C%20m%2Fs%7D%7B-%209.81%5C%20m%2Fs%5E2%7D%5C%5C)
<u>t = 1.53 s</u>
Now, for the height attained we will use the 2nd equation of motion:
![h = v_{i}t + \frac{1}{2}gt^2\\\\h = (15\ m/s)(1.53\ s) + \frac{1}{2}(9.81\ m/s^2)(1.53)^2\\\\h = 22.93\ m + 11.48\ m](https://tex.z-dn.net/?f=h%20%3D%20v_%7Bi%7Dt%20%2B%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%5C%5C%5C%5Ch%20%3D%20%2815%5C%20m%2Fs%29%281.53%5C%20s%29%20%2B%20%5Cfrac%7B1%7D%7B2%7D%289.81%5C%20m%2Fs%5E2%29%281.53%29%5E2%5C%5C%5C%5Ch%20%3D%2022.93%5C%20m%20%2B%2011.48%5C%20m)
<u>h = 34.41 m</u>
(b)
using the 3rd equation of motion for a height of 5 m:
![2gh = v_{f}^2 - v_{i}^2\\2(-9.81\ m/s^2)(5\ m) = (0\ m/s)^2 - v_{i}^2 \\v_{i} = \sqrt{98.1\ m^2/ s^2}\\](https://tex.z-dn.net/?f=2gh%20%3D%20v_%7Bf%7D%5E2%20-%20v_%7Bi%7D%5E2%5C%5C2%28-9.81%5C%20m%2Fs%5E2%29%285%5C%20m%29%20%3D%20%280%5C%20m%2Fs%29%5E2%20-%20v_%7Bi%7D%5E2%20%5C%5Cv_%7Bi%7D%20%3D%20%5Csqrt%7B98.1%5C%20m%5E2%2F%20s%5E2%7D%5C%5C)
<u>vi = 9.9 m/s</u>
<u></u>
(c)
At t = 1 s:
![v_{f1s} = v_{i} + gt\\v_{f1s} = 15\ m/s + (-9.81\ m/s^2)(1\ s)\\](https://tex.z-dn.net/?f=v_%7Bf1s%7D%20%3D%20v_%7Bi%7D%20%2B%20gt%5C%5Cv_%7Bf1s%7D%20%3D%2015%5C%20m%2Fs%20%2B%20%28-9.81%5C%20m%2Fs%5E2%29%281%5C%20s%29%5C%5C)
<u>vf_1s = 5.19 m/s</u>
<u></u>
<u></u>
<u>h_1s = 10.095 m</u>
<u></u>
At t = 4 s:
Since the ball covers the maximum height of 34.41 m in 1.53 s and then starts moving downward.
Therefore for the remaining 4 s - 1.53 s = 2.47 s, the initial velocity will be 0 m/s at the highest point and the value of g will be positive due to downward motion.
![v_{f4s} = v_{i} + gt\\v_{f4s} = 0\ m/s + (9.81\ m/s^2)(2.47\ s)\\](https://tex.z-dn.net/?f=v_%7Bf4s%7D%20%3D%20v_%7Bi%7D%20%2B%20gt%5C%5Cv_%7Bf4s%7D%20%3D%200%5C%20m%2Fs%20%2B%20%289.81%5C%20m%2Fs%5E2%29%282.47%5C%20s%29%5C%5C)
<u>vf_4s = 24.23 m/s</u>
<u></u>
<u></u>
now, for the position with respect to railing:
![h_{4s} = h - h_{down}\\h_{4s} = 34.41\ m - 29.92\ m\\](https://tex.z-dn.net/?f=h_%7B4s%7D%20%3D%20h%20-%20h_%7Bdown%7D%5C%5Ch_%7B4s%7D%20%3D%2034.41%5C%20m%20-%2029.92%5C%20m%5C%5C)
<u>h_4s = 4.49 m (below railing)</u>
<u></u>
Answer: You use friction when your bike stops.
Explanation: Force and friction affect our daily lives in numerous amounts of ways. For instance, when a football is kicked, it moves faster later after some time its force decreases due to friction. A common example of friciton is when a bike stops. When the brakes are applied the friction on the pads cause the bike to stop.