an 8 kilogram ball is fired horizontally form a 1 times 10^3 kilogram cannon initially at rest after having been fired the momen
2 answers:
After the ball is shot, the cannon will move at a speed of 2.4 m/s.
We must learn more about the rule of conservation of linear momentum in order to locate the solution.
How can I determine the cannon's post-ball velocity magnitude?
- According to the law of conservation of linear momentum, a system's starting and ultimate velocities are equal.
- Since the cannon's initial velocity is 0 in this case, the system's initial momentum will be equal to zero as well.
- We have,
- When a cannon fires a ball, the cannon will recoil quickly and in the opposite direction of the ball's motion.
- Given that it is now moving eastward, the recoil momentum is towards the west.
- As a result, when the ball is fired, the cannon's velocity will be,
Thus, we can infer that the cannon's maximum speed once the ball is fired is 2.4 m/s.
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The magnitude of the cannons velocity after the ball is fired will be 2.4m/s.
To find the answer, we have to know more about the law of conservation of linear momentum.
<h3>How to find the magnitude of the cannons velocity after the ball is fired?</h3>- The law of conservation of linear momentum states that, the initial momentum of a system will be equal to the final momentum.
- Here, then initial momentum of the system will be equal to zero, since the initial velocity of the cannon is equal to zero.
- Given that,
- When the ball is fired from the canon, then the cannon will have a recoil velocity in the opposite direction of motion of the ball.
- Since it has a final momentum towards east, the recoil momentum will be in the west.
- Thus, the velocity of the cannon after when the ball is fired will be,
Thus, we can conclude that, the magnitude of the cannons velocity after the ball is fired is 2.4m/s.
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directed downwards, and its initial vertical position at time t=0 is 750 m. So, the vertical position (in meters) at any time t can be written as
(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write
from which we find the time t after which the stone reaches the ground:
(c) The velocity of the stone at time t can be written as
because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
(d) if the stone has an initial velocity of
, then its law of motion would be
and we can find the time it needs to reach the ground by requiring again y(t)=0:
which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.
(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write
from which we find the time t after which the stone reaches the ground:
(c) The velocity of the stone at time t can be written as
because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
(d) if the stone has an initial velocity of
and we can find the time it needs to reach the ground by requiring again y(t)=0:
which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.
Answer:
40 J
Explanation: