Question

A projectile of mass 50kg is shot vertically upwards with an initial velocity of 100m/s. After 5s, it explodes into two fragments, one of which having mass 20kg travels vertically up with a velocity of 150m/s. What is the velocity of the other particile at that instant? Calculate the sum of momenta of the fragments 4s after explosion and show that it is equal to the momentum of the projectile at that instant if there were no explosion. (take g=9.8m/s2)

Answer 1

QUESTION:

A projectile of mass $\mathbf{m}_1 = 50\mathrm{kg}$ is shot vertically upwards with an initial velocity of $\mathbf{u}_0 = 100\mathrm{m / s}$ . After $\mathbf{t}_1 = 5\mathrm{s}$ , it explodes into two fragments, one of which having mass $\mathbf{m}_2 = 20\mathrm{kg}$ travels vertically up with a velocity of $\mathbf{u}_2 = 150\mathrm{m / s}$ . What is the velocity $\mathbf{u}_3$ of the other particle at that instant? Calculate the sum of momenta of the fragments $\mathbf{t}_2 = 4\mathrm{s}$ after explosion and show that it is equal to the momentum of the projectile at that instant if there were no explosion. (take $g = 9.8\mathrm{m / s2}$ )

SOLUTION:

(Attention - the length of the momentum vectors on this sketch isn't proportional to their modulus)

Let's find the velocity of the projectile in the moment of explosion:

$$v _ {1} = v _ {0} - g \cdot t _ {1} = 1 0 0 - 9. 8 \cdot 5 = 5 1 \mathrm {m} / \mathrm {s}$$

According to the momentum conservation law:

$$\begin{array}{l} \mathrm {p} _ {1} = \mathrm {p} _ {2} + \mathrm {p} _ {3} \\ \mathrm {m} _ {1} \mathrm {v} _ {1} = \mathrm {m} _ {2} \mathrm {v} _ {2} + \mathrm {m} _ {3} \mathrm {v} _ {3} \quad (\text {projection on y axis}) \\ \end{array}$$

Hence

$$v _ {3} = \frac {m _ {1} v _ {1} - m _ {2} v _ {2}}{m _ {3}} = \frac {5 0 \cdot 5 1 - 2 0 \cdot 1 5 0}{5 0 - 2 0} = - 1 5 m / s$$

(Here minus sign show, that velocity $\mathbf{v}_3$ is directed downward, and its projection on $y$ axis is negative. We can't predict the direction of $\mathbf{v}_3$ velocity without calculation and hence we assume first that it was directed upward, but calculations show that that velocity $\mathbf{v}_3$ is directed downward)

Let's find the velocities and momenta of the fragments after 4 s after explosion:

1st fragment :

$$\begin{array}{l} v _ {2} ^ {\prime} = v _ {2} - g t = 1 5 0 - 9. 8 \cdot 4 = 1 1 0. 8 \mathrm {m} / \mathrm {s} \\ p _ {2} ^ {\prime} = m _ {2} v _ {2} ^ {\prime} = 2 2 1 6 \mathrm {k g} \cdot \mathrm {m} / \mathrm {s} \\ \end{array}$$

2nd fragment

$$\begin{array}{l} v _ {3} ^ {\prime} = v _ {3} - g t = - 1 5 - 9. 8 \cdot 4 = - 5 4. 2 m / s \\ p _ {3} ^ {\prime} = m _ {3} v _ {3} ^ {\prime} = - 1 6 2 6 \mathrm {k g} \cdot \mathrm {m} / \mathrm {s} \\ p _ {2} ^ {\prime} + p _ {3} ^ {\prime} = 5 9 0 \mathrm {k g} \cdot \mathrm {m} / \mathrm {s} \\ \end{array}$$

Let's find now the velocity and momentum of projectile after $4s + 5s = 9s$ from shot, assuming that there was no explosion:

$$v _ {1} ^ {\prime} = v _ {0} - g \cdot \left(t _ {1} + t _ {2}\right) = 1 0 0 - 9. 8 = 1 1. 8 m / s$$ $$p _ {1} ^ {\prime} = m _ {1} v _ {1} ^ {\prime} = 5 9 0 \mathrm {k g} \cdot \mathrm {m} / \mathrm {s}$$

Hence

$$p ^ {\prime} = p _ {2} ^ {\prime} + p _ {3} ^ {\prime}$$

ANSWER

$$v _ {3} = - 1 5 \mathrm {m} / \mathrm {s}$$ $$p ^ {\prime} = p _ {2} ^ {\prime} + p _ {3} ^ {\prime} = 5 9 0 k g \cdot m / s$$