Question

A watermelon is blown into three pieces by a large firecracker. Two pieces of equal mass m fly away perpendicular to one another, one in the x direction another in the y direction. Both of these pieces fly away with a speed of V = 42 m/s. The third piece has three times the mass of the other two pieces. Randomized Variables V = 42 m/s show answer No Attempt 33% Part (a) Write an expression for the speed of the larger piece, that is in terms of only the variable V.

Answer 1

Answer:

Speed of larger piece is $\dfrac{V\sqrt{2}}{3}$

Explanation:

We apply the principle of conservation of momentum.

The watermelon is initially at rest. The initial momentum = 0 kg m/s in all directions.

After the collision,

Vertical momentum = momentum of piece in y-direction + y-component of momentum of larger piece = $mV + 3mv_{ly}$

Here, $v_{ly}$ is the y-component of velocity of larger piece.

This is equal to 0, since the initial momentum is 0.

$v_{ly}=\dfrac{V}{3}$

Horizontal momentum = momentum of piece in x-direction + x-component of momentum of larger piece = $mV + 3mv_{lx}$

Here, $v_{lx}$ is the x-component of velocity of larger piece.

This is also equal to 0, since the initial momentum is 0.

$v_{lx}=\dfrac{V}{3}$

The velocity of the larger piece, $v_l$, is the resultant of $v_{lx}$ and $v_{ly}$. Since they are mutually perpendicular,

$v_l = \sqrt{v_{ly}^2+v_{lx}^2}= \sqrt{\left(\dfrac{V}{3}\right)^2+\left(\dfrac{V}{3}\right)^2}$

$v_l = \dfrac{V\sqrt{2}}{3}$