Answer:
Answer B is the correct answer: "<em>Motion of one projectile as seen from the other is a straight line.</em>"
Explanation:
Let's write the equations of motion for each projectile, using that projectile 
is launched with velocity which has components associated with the angle of launching, given in x and y coordinates as: 
.
Similarly, assume that projectile b is launched with velocity 
with components due to the launching angle = 
then the equations of motion for the two projectiles launched at the same time (t) from the same spot (position that we assume to be at the origin of coordinates to simplify formulas) are:
therefore, from the frame of reference of projectile "b", the x and y position of projectile "" would be:
which is linear in "t"
![y_{a\,\,b}=y_a-y_b= a_y\,t-\frac{1}{2} g\,t^2-\left[ b_y\,t-\frac{1}{2} g\,t^2\right]=(a_y-b_y)\,t](https://tex.z-dn.net/?f=y_%7Ba%5C%2C%5C%2Cb%7D%3Dy_a-y_b%3D%20a_y%5C%2Ct-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2-%5Cleft%5B%20b_y%5C%2Ct-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2%5Cright%5D%3D%28a_y-b_y%29%5C%2Ct)
which is also linear in t.
Therefore the motion of one projectile with reference to the other is a straight line (answer B)
Notice as well that this two projectiles cannot collide because they have been launched together, and supposedly at different speeds and angles. The only way that they can share the same x-coordinate and the same y-coordinate at the same time "t" is if their velocity components are equal, which is not what we are told.
(a) The momentum of the first trolley is 5.4 kgm/s
(b) The velocity of the trolleys after impact is 2.7m/s
<u>Explanation:</u>
Given:
Mass, m₁ = 1.2kg
Velocity, v₁ = 4.5m/s
Mass, m₂ = 0.8kg
v₂ = 0
(a) Momentum of the trolley before impact, p
We know:
Momentum = mass X velocity
p = 1.2 X 4.5
p = 5.4 kgm/s
Therefore, the momentum of the first trolley is 5.4 kgm/s
(b) Speed of the trolleys after impact, v = ?
During collision, the momentum is conserved.
So,
m₁v₁ + m₂v₂ = (m₁ + m₂)v
(1.2 X 4.5) + (0.8 X 0) = (1.2 +0.8) X v
5.4 + 0 = 2v
v = 2.7m/s
Therefore, the velocity of the trolleys after impact is 2.7m/s
Answer:
Force of the ball against the bat
Explanation:
Newton's third law of motion states that for every action, there is an equal and opposite reaction. When a player hits a ball with a bat, the force the player exerted on the ball through the bat gave the ball a push force in the forward direction. This is an action force. Now that an action has been initiated, there must be a reaction according to Newton's third law of motion. The magnitude of the force on the ball equals the magnitude of the force on the bat. The direction of the force on the ball is opposite to the direction of the force on the ball. The ball will give the bat an equal and opposite force trying to push the bat backward by resisting the impact of the bat. This is the reaction. The ball will only move if the applied force is able to overcome the resisting force by the bat.
Where;
is the action force of the bat against the ball.
is the equal and opposite reaction force of the ball towards the ball.
change in velocity is 3m/s,the force exterted by bat on ball is 4.8N and the acceleration is 30m/s².
<h3>Give some difference between velocity and speed.</h3>
1) Speed is scalar quantity but velocity is vector quantity.
2) Speed is distance followed by an individual with respect to time and velocity is displacement with respect to time .
1) final velocity- intial velocity
25-22= 3m/s
2) The force is
F= ma
. 0.16×30= 4.8N
to learn more about speed and velocity click brainly.com/question/19425181
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