Answer:
261.3 m/s
Explanation:
Mass of bullet=m=15 g=
1 kg=1000g
Mass of block=M=3 kg
d=0.086 m
Total mass =M+m=3+0.015=3.015 kg
K.E at the time strike=Gravitational potential energy at the end of swing
Using g=
Substitute the values
Velocity after collision=V=1.3 m/s
Velocity of block=v'=0
Using conservation law of momentum
Using the formula
what? I guess:
- practice different habits. If you fail don't give up.
- don't always trust people, some are not what they seem.
this question doesn't make any sense...
<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>
The sum of potential energy<span> and kinetic </span><span>energy.
Hope I helped!</span>
