Question

1. a frustrated physics student blew up her physics textbook, using a small amount of an explosive. it broke into three pieces, which miraculously flew off in directions that were all in the same geometric plane. a 0.200 kg piece flew off at 20.0 m/s, and a 0.100 kg piece went off at 90° to the first piece, at 30.0 m/s. (a) what was the momentum of the third piece? (b) if the mass of the third piece was 0.150 kg, what was its velocity right after the explosion

Explanation:

Step1: Recall momentum - conservation principle

In an explosion, the total initial momentum is zero (assuming the textbook was at rest initially). According to the law of conservation of momentum, the total final momentum is also zero. Let the momentum of the first piece be $\vec{p}_1$, the second piece be $\vec{p}_2$ and the third piece be $\vec{p}_3$. So, $\vec{p}_1+\vec{p}_2+\vec{p}_3 = 0$.

Step2: Calculate the momentum of the first piece

The mass of the first piece $m_1 = 0.200$ kg and its velocity $v_1=20.0$ m/s. The momentum of the first piece $p_1=m_1v_1=(0.200\ kg)\times(20.0\ m/s) = 4.00\ kg\cdot m/s$.

Step3: Calculate the momentum of the second piece

The mass of the second piece $m_2 = 0.100$ kg and its velocity $v_2 = 30.0$ m/s. The momentum of the second piece $p_2=m_2v_2=(0.100\ kg)\times(30.0\ m/s)=3.00\ kg\cdot m/s$.

Step4: Use vector - addition of momentum

Since the first and second pieces fly off at right - angles to each other, we can use the Pythagorean theorem to find the magnitude of the combined momentum of the first two pieces. $p_{12}=\sqrt{p_1^{2}+p_2^{2}}=\sqrt{(4.00\ kg\cdot m/s)^{2}+(3.00\ kg\cdot m/s)^{2}}=\sqrt{16 + 9}\ kg\cdot m/s=\sqrt{25}\ kg\cdot m/s = 5.00\ kg\cdot m/s$.

Step5: Find the momentum of the third piece

Since $\vec{p}_1+\vec{p}_2+\vec{p}_3 = 0$, the magnitude of the momentum of the third piece $p_3$ is equal to the magnitude of the combined momentum of the first two pieces, so $p_3 = 5.00\ kg\cdot m/s$.

Step6: Calculate the velocity of the third piece

The mass of the third piece $m_3=0.150$ kg. Using the formula $p = mv$, we can find the velocity $v_3$ of the third piece. $v_3=\frac{p_3}{m_3}=\frac{5.00\ kg\cdot m/s}{0.150\ kg}=\frac{100}{3}\ m/s\approx33.3\ m/s$.

Answer:

(a) The momentum of the third piece is $5.00\ kg\cdot m/s$.
(b) The velocity of the third piece is $\frac{100}{3}\ m/s\approx33.3\ m/s$.