Answer:
Part a)
v = 16.52 m/s
Part b)
v = 7.47 m/s
Explanation:
Part a)
(a) when the large-mass object is the one moving initially
So here we can use momentum conservation as the net force on the system of two masses will be zero
so here we can say
since this is a perfect inelastic collision so after collision both balls will move together with same speed
so here we can say
Part b)
(b) when the small-mass object is the one moving initially
here also we can use momentum conservation as the net force on the system of two masses will be zero
so here we can say
Again this is a perfect inelastic collision so after collision both balls will move together with same speed
so here we can say
Force = mass times acceleration
F = 21000 x 36.9 = 774900
Therefore, 774900N force is required.
<span> For any body to move in a circle it requires the centripetal force (mv^2)/r.
In this case a ball is moving in a vertical circle swung by a mass less cord.
At the top of its arc if we draw its free body diagram and equate the forces in radial
direction to the centripetal force we get it as T +mg =(mv^2)/r
T is tension in cord
m is mass of ball
r is length of cord (radius of the vertical circle)
To get the minimum value of velocity the LHS should be minimum. This is possible when T = 0. So
minimum speed of ball v at top =sqrtr(rg)=sqrt(1.1*9.81) = 3.285 m/s
In the second case the speed of ball at top = (2*3.285) =6.57 m/s
Let us take the lowest point of the vertical circle as reference for potential energy and apllying the conservation of energy equation between top & bottom
we get velocity at bottom as 9.3m/s.
Now by drawing the free body diagram of the ball at the bottom and equating the net radial force to the centripetal force
T-mg=(mv^2)/r
We get tension in cord T=13.27 N</span>
we have to use newtons law of gravitation which is
F=GMm/r^2
G=6.67 x 10^<span>-11N kg^2/m^2
</span>M=<span>(15kg)
</span>m=15 kg
r=(3.0m)^2<span>
</span>putting values we have
<span>=(6.67 x 10^-11N kg^2/m^2)(15kg)(15kg)/(3.0m)^2 </span>
=1.67 x 10^-9N
