Question

A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular to each other, at 17 m/s. The third piece has twice the mass as the other two.Part 1
What is the speed of the third piece?Part 2
What is the direction of the third piece?

Answer 1

Answer:

12.02 m/s north of east

Explanation:

Let $v_x, v_y$ be the horizontal and vertical velocity of the 3rd piece, positive in the East and North direction. Also let m be the mass of the 2 smaller pieces, so 2m is the mass of the 3rd piece.

We can solve for the 2 velocities components using law of momentum conservation. Before the explosion, the total momentum was 0 because the coconut was at rest. After the explosion:

- In the x direction:  

$2mv_x + m*(-17) = 0$ negative 17 because 1 piece flew west

$v_x = 17/2 = 8.5 m/s$

- In the y direction

$2mv_y + m(-17) = 0$ negative 17 because 1 piece flew south

$v_y = 17/2 = 8.5 m/s$

So the speed and direction of the 3rd piece is

$v = \sqrt{v_y^2 + v_x^2} = \sqrt{8.5^2 + 8.5^2} = \sqrt{72.25 + 72.25} = \sqrt{144.5} = 12.02 m/s$

$tan\theta = \frac{v_y}{v_x} = \frac{8.5}{8.5} = 1$

$\theta = tan^{-1}1 = 0.79 rad \approx 45^o$ north of east

Answer 2

Answer:

v3 = –17m/s East or north.

Explanation:

This problem involves the concept of momentum

The total momentum before the explosion of the fire cracker is zero

Let the mass of the first two particles be m1 = m2 = m, velocities v1 = v2 = 17m/s

And m3 = 2m

Taking south and west as positive directions, total momentum before collision equals total momentum after collision

Summing momentum along the north-south axis

0 = m1v1y +m2v2y+ m3v3y

0 = m×17 +m×0 + 2m×v3

0 = 17m + 2mv3y

Dividing through by m

0= 17 +2v3y

v3y = –17/2 = –8.5m/s

0 = m1v1x +m2v2x + m3v3x

0 = m×0 + m×17 + 2m×v3x

0 = 17m + 2mv3x

Dividing through by m

0= 17 +2v3x

v3y = –17/2 = – 8.5m/s

V3 = √(v3x² + v3y²)

V3 = √((-8.5)² + (-8.5)²)

V3 = √(144.5) = 12.02m/s

θ = Tan -¹(v3y/v3x) = Tan -¹(–8.5/–8.5) = 45°

θ = 45° (NE)