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 = 25 m/s. The third piece has three times the mass of the other two pieces.
a. Write an expression for the speed of the larger piece, that is in terms of only the variable V.
b. What is the numeric value for the speed of the larger piece, in meters per second?
c. At what angle does the largest piece travel with respect to the -y axis, in degrees?

Answer 1

Answer:

a) $v_{3}=\frac{V\sqrt{2}}{3}$

b) $v_{3}=11.79 m/s$

c) $\alpha=45^{\circ}$

Explanation:

a) Let's use the conservation of momentum, it means the change of momentum in the system is equal to zero.

The inial total momentum is zero and the final total momentum is the sum of each particle momentum.

$0=m_{1}v_{1}+m_{2}v_{2}+m_{3}v_{3}$

• m(1) is the mass of the piece in x direction
• m(2) is the mass of the piece in y direction
• m(3) is the mass of the piece in some direction with a angle with respect to -y.          

But we know that $m_{1}=m_{2$ and $v_{1}=v_{2}=V$ and $m_{3}=3m_{1}$, so we have the vectorial equation:

$0=m_{1}V\hat{i}+m_{1}V\hat{j}+3m_{1}\vec{v_{3}}$

Now we can find the vector v₃ in terms of V.

$\vec{v_{3}}=-\frac{V}{3}(\hat{i}^{2}+\hat{j}^{2})$    

Let's find now the magnitude of this vector:

$v_{3}=\frac{V}{3}\sqrt{(\hat{-i})^{2}+(\hat{-j})^{2}}=\frac{V\sqrt{2}}{3}$

b) We just need to replace V in the above equation;

$v_{3}=\frac{25\sqrt{2}}{3}=11.79 m/s$

c) The vector velocity of the particle 3 is in term of components, x, and y. So we can use tangent to find the angle.

$tan(\alpha)=\frac{V_{x}}{V_{y}}$

but we know that Vx and Vy are the same so we have:

$tan(\alpha)=1$

$\alpha=arctan(1)=45^{\circ}$

       

I hope it helps you!