Question

A bomb at rest at the origin of an xy-coordinate system explodes into three pieces. Just after the explosion, one piece, of mass 5 kg, moves with a velocity of 42 m/s in the negative x-direction, and a second piece, also of mass 5 kg, moves with a velocity of 38 m/s in the negative y-direction. The third piece has a mass of 10 kg.a. What is the initial momentum of the bomb before the explosion?b. What is the x-component of the velocity of the third piece just after the explosion?c. What is the y-component of the velocity of the third piece just after the explosion?d. What is the magnitude of the velocity of the third piece?e. Just after the explosion, in what direction is the third piece moving? (Give your answer as an angle measured clockwise from the positive x-axis.)

Answer 1

Answer:

Explanation:

a ) It is given that bomb was at rest initially , so ,  its momentum before the explosion was zero.

b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.

If v be the velocity of the third part along a direction making angle θ

with x axis ,

x component of v = vcosθ

So momentum along x axis  after explosion of third part   = mv cosθ

= 10 v cosθ

Momentum along x of first part = -  5 x 42 m/s

momentum of second part along x direction =0

total momentum along x direction before explosion = total momentum along x direction after explosion

0 = - 5 x 42 + 10 v cosθ

v cosθ = 21

Similarly

total momentum along y direction before explosion = total momentum along y direction after explosion

0 = - 5 x 38 +  10 v sinθ

v sinθ= 21

squaring and and then adding the above equation

v² cos²θ +v² sin²θ = 21² +19²

v² = 441 + 361

v = 28.31 m/s

Tanθ = 21 / 19

θ = 48°