Answer:
The heavier piece acquired 2800 J kinetic energy
Explanation:
From the principle of conservation of linear momentum:
0 = M₁v₁ - M₂v₂
M₁v₁ = M₂v₂
let the second piece be the heavier mass, then
M₁v₁ = (2M₁)v₂
v₁ = 2v₂ and v₂ = ¹/₂ v₁
From the principle of conservation of kinetic energy:
¹/₂ K.E₁ + ¹/₂ K.E₂ = 8400 J
¹/₂ M₁(v₁)² + ¹/₂ (2M₁)(¹/₂v₁)² = 8400
¹/₂ M₁(v₁)² + ¹/₄M₁(v₁)² = 8400
K.E₁ + ¹/₂K.E₁ = 8400
Now, we determine K.E₁ and note that K.E₂ = ¹/₂K.E₁
1.5 K.E₁ = 8400
K.E₁ = 8400/1.5
K.E₁ = 5600 J
K.E₂ = ¹/₂K.E₁ = 0.5*5600 J = 2800 J
Therefore, the heavier piece acquired 2800 J kinetic energy
The velocity of the submarine immediately after firing the missile is 0.0104 m/s
Explanation:
Mass of the submarine M=50 tonne=
Mass of the missile m=40 kg
velocity of the missile v= 13m/s
we have to calculate the velocity of the submarine after firing
This is the recoil velocity and its expression is derived from the law of conservation of momentum
recoil velocity of the submarine
Answer:
Explanation:
The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by
where
is the horizontal velocity
t is the time of flight
The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.
In the first situation, the horizontal distance covered is
in the second case, the horizontal velocity is increased to
And so the new distance travelled will be
So, the distance increases linearly with the horizontal velocity.