Which one? A? B? C? D?
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4. The Coyote has an initial position vector of .
4a. The Coyote has an initial velocity vector of . His position at time
is given by the vector
where is the Coyote's acceleration vector at time
. He experiences acceleration only in the downward direction because of gravity, and in particular
where
. Splitting up the position vector into components, we have
with
The Coyote hits the ground when :
4b. Here we evaluate at the time found in (4a).
5. The shell has initial position vector , and we're told that after some time the bullet (now separated from the shell) has a position of
.
5a. The vertical component of the shell's position vector is
We find the shell hits the ground at
5b. The horizontal component of the bullet's position vector is
where is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for
:
Answer:
46.48m/s
Explanation:
The problem is a combination of the principle of conservation of linear momentum and projectile motion.
The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.
For two bodies of masses and
moving with velocities
and
before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);
were v is their common velocity after impact. If the second mass was at rest before the impact, then its initial velocity
. therefore
. Equation (1) then becomes;
In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.
Given;
Substituting these values into (2), we get the following;
Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;
were t is the time taken to fall through the height h. To obtain t we use the second equation of free fall under gravity;
were g is acceleration due to gravity taken as . Therefore;
We then substitute R and t into equation (4) to obtain v.
We now further substitute this value of v into (3) to obtain ;