A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by υ⃗ =[5.00m/
1 answer:
You might be interested in
Answer:
(a) The maximum height achieved by the first ball, m₁ is 0.11 m
(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m
Explanation:
Given;
mass of the first ball, m₁ = 1 kg
mass of the second ball, m₂ = 2 kg
The velocity of the first when released from a height of 1 m before collision;
u₁² = u₀² + 2gh
u₀ = 0, since it was released from rest
u₁² = 2gh
u₁² = 2 x 9.8 x 1
u₁² = 19.6
u₁ = √19.6
u₁ = 4.427 m/s
The velocity of the second ball before collision, u₂ = 0
Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
where;
v₁ is the final velocity of the first ball after an elastic collision
v₂ is the final velocity of the second ball after an elastic collision
m₁u₁ + m₂(0) = m₁v₁ + m₂v₂
m₁u₁ = m₁v₁ + m₂v₂
1 x 4.427 = v₁ + 2v₂
v₁ + 2v₂ = 4.427
v₁ = 4.427 - 2v₂ ----- equation (1)
one directional velocity;
u₁ + v₁ = u₂ + v₂
u₂ = 0
u₁ + v₁ = v₂
v₁ = v₂ - u₁
v₁ = v₂ - 4.427 ------ equation (2)
Substitute v₁ into equation (1)
v₂ - 4.427 = 4.427 - 2v₂
3v₂ = 4.427 + 4.427
3v₂ = 8.854
v₂ = 8.854 / 3
v₂ = 2.95 m/s (→ forward direction)
v₁ = v₂ - 4.427
v₁ = 2.95 - 4.427
v₁ = - 1.477 m/s
v₁ = 1.477 m/s ( ← backward direction)
Apply the law of conservation of mechanical energy
(a) The maximum height achieved by the first ball (v₁ = 1.477 m/s)
(b) The maximum height achieved by the second ball (v₂ = 2.95 m/s)
Answer:
4.9 m/s²
Explanation:
Draw a free body diagram. There are two forces on the object:
Weight force mg pulling straight down,
and normal force N pushing perpendicular to the plane.
Sum the forces in the parallel direction.
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 30°)
a = 4.9 m/s²
