A 1.35-kg block of wood sits at the edge of a table, 0.730 m above the floor. A 1.15×10−2-kg bullet moving horizontally with a speed of 745 m/s embeds itself within the block.What horizontal distance does the block cover before hitting the ground?
Gravitational acceleration = g = -9.81 m/s2
Mass of the bullet = m1 = 1.15 x 10-2 kg
Mass of the block of wood = m2 = 1.35 kg
Initial velocity of the bullet = V1 = 745 m/s
Initial velocity of the block of wood = V2 = 0 m/s
Velocity of the block of wood and bullet after the collision = V3
By conservation of linear momentum,
m1V1 + m2V2 = (m1 + m2)V3
(1.15x10-2)(745) + (1.35)(0) = (1.15x10-2 + 1.35)V3
V3 = 6.293 m/s
The block will leave the table horizontally therefore,
Horizontal velocity of the block when it leaves the table = V3x = 6.293 m/s
Vertical velocity of the block when it leaves the table = V3y = 0 m/s
Height of the table = H = 0.73 m
Time taken by the block to hit the floor = T
The displacement of the block in the vertical direction when it hits the floor is downwards therefore it is negative.
-H = V3yT + gT2/2
-0.73 = (0)T + (-9.81)T2/2
T = 0.386 sec
Horizontal distance covered by the block = R
There is no horizontal force acting on the block therefore the horizontal velocity of the block remains constant.
R = V3xT
R = (6.293)(0.386)
R = 2.43 m
Horizontal distance the block covers before hitting the ground = 2.43 m
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