Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of launch is 75.0° and 15.0° resp
2 answers:
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Answer:
Therefore,
Final velocity of the coupled carts after the collision is
Explanation:
Given:
Mass of Glidding Cart = m₁ = 2 kg
Mass of Stationary Cart = m₂ = 5 kg
Initial velocity of Glidding Cart = u₁ = 14 m/s
Initial velocity of Stationary Cart = u₂ = 0 m/s
To Find:
Final velocity of the coupled carts after the collision =
Solution:
Law of Conservation of Momentum:
For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.
It is denoted by "p" and given by
Momentum = p = mass × velocity
Hence by law of Conservation of Momentum we hame
Momentum before collision = Momentum after collision
Here after collision both are stuck together so both will have same final velocity,
Substituting the values we get
Therefore,
Final velocity of the coupled carts after the collision is
Answer:
v = 79.2 m/s
Solution:
As per the question:
Mass of the object, m = 250 g = 0.250 kg
Angle,
Coefficient of kinetic friction,
Mass attached to the string, m = 0.200 kg
Distance, d = 30 cm = 0.03 m
Now,
The tension in the string is given by:
(1)
Also
T = m(g + a)
Thus eqn (1) can be written as:
Now, the speed is given by the third eqn of motion with initial velocity being zero:
where
u = initial velocity = 0
Thus
