Question

An object is tracked by a radar station and found to have a position vector by r=(3500-160t)i + 2700j + 300k, with r in m and t in sec. The radar station's x-axis points east, its y-axis north and its z-axis vertically up. If the object is a 250-kg missile warhead, what are (a) its linear momentum (b) its direction of motion (c) its net force on it.

Answer 1

Answer on Question #40446

Physics - Mechanics | Kinematics | Dynamics

Question:

An object is tracked by a radar station and found to have a position vector by $r = (3500 - 160t)i + 2700j + 300k$, with $r$ in $m$ and $t$ in sec. The radar station's x-axis points east, its y-axis north and its z-axis vertically up. If the object is a 250-kg missile warhead, what are (a) its linear momentum (b) its direction of motion (c) its net force on it.

Solution:

The linear momentum is

$$\vec {p} = m \vec {v},$$

where $m = 250 \, \text{kg}$ is mass of the object and $v$ is its velocity.

$$\vec {v} = \frac {d \vec {r}}{d t} = - 160 \vec {i}.$$

Thus,

$$\vec {p} = - 160 m \vec {i} = - 40,000 \vec {i} \frac {k g \cdot m}{s}.$$

Because only the X coordinate of the object is changed in time and X axis points East, the object moves to the East.

The net force $F_{net} = ma$ acting on the object is zero, because the acceleration

$$a = \frac {d \vec {v}}{d t} = \frac {d ^ {2} \vec {r}}{d t ^ {2}} = 0.$$

Answer:

a) $\vec{p} = -40,000\vec{i}\frac{\mathrm{kg}\cdot\mathrm{m}}{\mathrm{s}}$

b) East

c) 0.