A block of mass 1 . 9   kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0 . 1   kg collides with the block and sticks to it. If the velocity of the bullet is 20   m   s - 1 in the horizontal direction just before the collision then the kinetic energy just…

Question

A block of mass 1 . 9 &#160; kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0 . 1 &#160; kg collides with the block and sticks to it. If the velocity of the bullet is 20 &#160; m &#160; s - 1 in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take g &#160; = &#160; 10 &#160; m &#160; s - 2 . Assume there is no rotational motion and loss of energy after the collision is negligible.]

Aicharya · Accepted Answer

Conservation of linear momentum 0 . 1 &#215; 20 = 0 . 1 + 1 . 9 &#215; v v = 1 &#160; m &#160; s - 1 Using work energy theorem W g = &#8710; k 2 &#215; g &#215; 1 = k - 1 2 &#215; 2 &#215; 1 2 &#8756; &#160; &#160; &#160; k = 21 &#160; J

A block of mass 1 . 9   kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0 . 1…

Solution

A block of mass 1 . 9 &#160; kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0 . 1…

Solution

A block of mass 1 . 9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0 . 1…