A 2.8-kg physics cart is moving forward with a speed of 45 cm/s. A 1.9-kg brick is dropped from rest and lands on the cart. The
1 answer:
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Explanation:
सिद्ध कीजिए किसी भी बराबर भुजाओं वाले त्रिभुज में उनके सामने के कोण बराबर होते है
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:
Force on the proton will be
Explanation:
We have given speed of proton is 20.3% of speed of light
Speed of light
So speed of proton
Magnetic field B = 0.00629 T
Charge on proton
Angle between velocity and magnetic field
Force on the proton is equal to