By law of conservation of momentum:
Summation of momentum before collision = Summation of momentum after collision
Momentum = mass*velocity
If v and u represents initial and final velocities and assuming that the bred marble was at rest before collision;
m1u1+m2u2 = m1v1+m2v2
But u2 = 0
Then,
m1u1 = m1v1+m2v2
u1 = (m1v1+m2v2)/m1
Where;
u1 = Initial velocity of red marble before collision
m1 = mass of blue marble = 0.1 kg
v1 = Final velocity of red marble = 10 m/s
v2 = Final velocity of blue marble = 15 m/s
m2 = mass of blue marble = 70 g = 0.07 kg
Substituting;
u1 = (0.1*10+0.07*15)/0.1 = 20.5 m/s
Therefore, red marble original velocity before collision was 20.5 m/s.
Answer:
<h2>4</h2>
Explanation:
Mechanical advantage is defined as the ratio of the load to the effort applied to raise the load. If minimal effort is used to overcome a much larger load, that is when we have what is called mechanical advantage i.e a machine has been used to our advantage.
Mathematically, MA = Load/Effort
Given parameters
Load = 160N
Effort = 40N
Required
Mechanical Advantage
Using the formula above
MA = 160N/40N
MA = 4
<em>advantage</em>
Answer:
(a) Amplitude=0.0760 m
(b) Speed=0.337 m/s
Explanation:
(a) For amplitude
We can use the mentioned description of the motion and the energy conservation principle to find amplitude of oscillatory motion
![k_{i}+U_{i}=K_{f}+U_{f}\\ (1/2)mv^{2}+0=0+(1/2)kA^{2}\\ A^{2}=\frac{mv^{2}}{k} \\A=\sqrt{\frac{mv^{2}}{k}}\\ A=\sqrt{\frac{m}{k} }v\\ A=\sqrt{\frac{(0.800kg)}{16N/m} }(0.34m/s)\\A=0.0760m](https://tex.z-dn.net/?f=k_%7Bi%7D%2BU_%7Bi%7D%3DK_%7Bf%7D%2BU_%7Bf%7D%5C%5C%20%281%2F2%29mv%5E%7B2%7D%2B0%3D0%2B%281%2F2%29kA%5E%7B2%7D%5C%5C%20%20A%5E%7B2%7D%3D%5Cfrac%7Bmv%5E%7B2%7D%7D%7Bk%7D%20%5C%5CA%3D%5Csqrt%7B%5Cfrac%7Bmv%5E%7B2%7D%7D%7Bk%7D%7D%5C%5C%20A%3D%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%20%7Dv%5C%5C%20A%3D%5Csqrt%7B%5Cfrac%7B%280.800kg%29%7D%7B16N%2Fm%7D%20%7D%280.34m%2Fs%29%5C%5CA%3D0.0760m)
(b) For Speed
Again we can use the mentioned description of the motion and the energy conservation principle to find amplitude of oscillatory motion
![k_{i}+U_{i}=K_{f}+U_{f}\\ (1/2)m(v_{i})^{2}+0=(1/2)m(v_{f} )^{2}+(1/2)k(A/2)^{2}\\ (1/2)m(v_{i})^{2}=(1/2)m(v_{f} )^{2}+(1/2)k(A/2)^{2}\\(1/2)m(v_{i})^{2}-(1/2)k(A/2)^{2}=(1/2)m(v_{f} )^{2}\\(1/2)[m(v_{i})^{2}-k(A/2)^{2}]=(1/2)m(v_{f} )^{2}\\(v_{f} )^{2}=1/m[m(v_{i})^{2}-k(A/2)^{2}]\\As\\x=0.250A\\(v_{f} )^{2}=(1/0.800kg)[0.800kg(0.34m/s)^{2}-(16N/m)(0.250(0.07602m)/2)^{2}\\(v_{f} )^{2}=0.1138\\ v_{f}=\sqrt{0.1138}\\ v_{f}=0.337m/s](https://tex.z-dn.net/?f=k_%7Bi%7D%2BU_%7Bi%7D%3DK_%7Bf%7D%2BU_%7Bf%7D%5C%5C%20%281%2F2%29m%28v_%7Bi%7D%29%5E%7B2%7D%2B0%3D%281%2F2%29m%28v_%7Bf%7D%20%29%5E%7B2%7D%2B%281%2F2%29k%28A%2F2%29%5E%7B2%7D%5C%5C%20%281%2F2%29m%28v_%7Bi%7D%29%5E%7B2%7D%3D%281%2F2%29m%28v_%7Bf%7D%20%29%5E%7B2%7D%2B%281%2F2%29k%28A%2F2%29%5E%7B2%7D%5C%5C%281%2F2%29m%28v_%7Bi%7D%29%5E%7B2%7D-%281%2F2%29k%28A%2F2%29%5E%7B2%7D%3D%281%2F2%29m%28v_%7Bf%7D%20%29%5E%7B2%7D%5C%5C%281%2F2%29%5Bm%28v_%7Bi%7D%29%5E%7B2%7D-k%28A%2F2%29%5E%7B2%7D%5D%3D%281%2F2%29m%28v_%7Bf%7D%20%29%5E%7B2%7D%5C%5C%28v_%7Bf%7D%20%29%5E%7B2%7D%3D1%2Fm%5Bm%28v_%7Bi%7D%29%5E%7B2%7D-k%28A%2F2%29%5E%7B2%7D%5D%5C%5CAs%5C%5Cx%3D0.250A%5C%5C%28v_%7Bf%7D%20%29%5E%7B2%7D%3D%281%2F0.800kg%29%5B0.800kg%280.34m%2Fs%29%5E%7B2%7D-%2816N%2Fm%29%280.250%280.07602m%29%2F2%29%5E%7B2%7D%5C%5C%28v_%7Bf%7D%20%29%5E%7B2%7D%3D0.1138%5C%5C%20v_%7Bf%7D%3D%5Csqrt%7B0.1138%7D%5C%5C%20v_%7Bf%7D%3D0.337m%2Fs)