A bullet of mass 50 g is fired from below into the bob of mass 450 g o

1.	A bullet of mass 50 g is fired from below into the bob of mass 450 g of a long simple pendulum as shown in figure. The bullet remains inside the bob and the bob rises through a height of 1.8 m. Find the speed of the bullet. Take g = 10 m/s2HOW CAN WE APPLY CONSERVATION OF MOMENTUM IN THIS QUESTION DURING THE COLLISON AS AN EXTERNAL FORCE ( DUE TO GRAVITY) IS ALSO ACTING ?
Answer» Explanation: Let the speed of the bullet be V. Let the common velocity of the bullet and the bob, after the bullet is EMBEDDED into the bob, is V. By the principle of conservation of the LINEAR momentum, V=0.45kg+0.05kg(0.05kg)v=10v The string becomes loose and the bob will go up with an ACCELERATION of g=9.8m/s2. As it comes to rest at a height of 1.8 m, using the EQUATION v2=u2+2ax, 1.8m=2×10m/s2(v/10)2 or v=60m/s.

A bullet of mass 50 g is fired from below into the bob of mass 450 g of a long simple pendulum as shown in figure. The bullet remains inside the bob and the bob rises through a height of 1.8 m. Find the speed of the bullet. Take g = 10 m/s2HOW CAN WE APPLY CONSERVATION OF MOMENTUM IN THIS QUESTION DURING THE COLLISON AS AN EXTERNAL FORCE ( DUE TO GRAVITY) IS ALSO ACTING ?

Answer»

Explanation:

Let the speed of the bullet be V. Let the common velocity of the bullet and the bob, after the bullet is EMBEDDED into the bob, is V. By the principle of conservation of the LINEAR momentum,

V=0.45kg+0.05kg(0.05kg)v=10v

The string becomes loose and the bob will go up with an ACCELERATION of g=9.8m/s2. As it comes to rest at a height of 1.8 m, using the EQUATION v2=u2+2ax,

1.8m=2×10m/s2(v/10)2

or v=60m/s.

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