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Grade: 11
        
A man stands still on a large sheet of slick ice; in his hand he holds a lighted firecracker. He throws the firecracker at an angle (that is, not vertically) into the air. Describe briefly, but as exactly as you can, the motion of the center of mass of the firecracker and the motion of the center of mass of the system consisting of man and firecracker. It will be most convenient to describe each motion during each of the following periods: (a) after he throws the firecracker, but before it explodes; (b) between the explosion and the first piece of firecracker hitting the ice; (c) between the first fragment hitting the ice and the last fragment landing; and (d) during the time when all fragments have landed but none has reached the edge of the ice.
5 years ago

Answers : (1)

Kevin Nash
askIITians Faculty
332 Points
							234-58_1.PNG
Therefore, the force experienced by the man is equal to the force experienced by the firecracker. However, due to larger inertia, the acceleration possessed by the man would be smaller than the acceleration possessed by the firecracker. From, this one can deduce that the horizontal displacement of the firecracker (from the point it was thrown) will be greater than the man corresponding to same time, and the horizontal position of the center of mass is always to the right of the point from which it was released
The vertical force experienced by the man is directed downwards, and for the firecracker is directed upwards. Therefore, the firecracker will displace vertically but the man does not. If H represents the height of the firecracker after time t , then the vertical position
of center of mass is \frac{H}{2}.
(b)
Assume that the man came to rest before the firecracker explodes into pieces. The firecracker breaks due to internal forces, and therefore the linear momentum of the firecracker before it exploded will be the same the total linear momentum of pieces after explosion. From this one can derive that the center of mass of the pieces of firecracker will follow the path of the original unbroken firecracker.
Therefore, the center of mass of the system will be move in the same way as in part (a), except the fact that the center of mass will shift more quickly towards the right. This is because the motion of the man has ceased down while the motion of the firecracker still continues.
(c)
It is important to note that to simplify the problem and avoid the complexity creeping inn due to the splitting into pieces; one should replace the motion of pieces of firecracker with the motion of original firecracker. From the height of the firecracker at any instant one can calculate the vertical position of the center of mass of system together with the horizontal position.
(d)
When all the fragments lands on the ice sheet, they move horizontally towards its edge. Therefore the vertical position of the center of mass of the system can be neglected (ycm = 0) , while the horizontal position of center of mass if given by the formula, 234-1019_1.PNG
5 years ago
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Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


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