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Shalnov [3]
3 years ago
9

A child on a 2.4 kg scooter at rest throws a 2.2 kg ball. The ball is given a speed of 3.1 m/s and the child and scooter move in

the opposite directions at 0.45 m/s. Find the child's mass.
Physics
1 answer:
kykrilka [37]3 years ago
3 0

Answer:

The child's mass is 14.133 kg

Explanation:

From the principle of conservation of linear momentum, we have;

(m₁ + m₂) × v₁ + m₃ × v₂ = (m₁ + m₂)  × v₃ - m₃ × v₄

We include the negative sign as the velocities were given as moving in the opposite directions

Since the child and the ball are at rest, we have;

v₁ = 0 m/s and v₂= 0 m/s

Hence;

0 = m₁ × v₃ - m₂ × v₄

(m₁ + m₂)× v₃ = m₃ × v₄

Where:

m₁ = Mass of the child

m₂ = Mass of the scooter = 2.4 kg

v₃ = Final velocity of the child and scooter = 0.45 m/s

m₃ = Mass of the ball = 2.4 kg

v₄ = Final velocity of the ball = 3.1 m/s

Plugging the values gives;

(m₁ + 2.4)× 0.45 = 2.4 × 3.1

(m₁ + 2.4) = 16.533

∴ m₁ + 2.4 = 16.533

m₁ = 16.533 - 2.4 = 14.133 kg

The child's mass = 14.133 kg.

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Which of the following are found within the electromagnetic spectrum? Check all that apply. sound waves visible light X rays ult
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8 0
3 years ago
Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts
olga2289 [7]

Answer:

a)  a = - 0.106 m/s^2  (←)

b) T = 12215.1064 N

Explanation:

If

F₁ = 9*1350 N = 12150 N   (→)

F₂ = 9*1365 N = 12285 N  (←)

∑Fx = M*a = (M₁  +M₂)*a           (→)

F₁ - F₂ = (M₁  +M₂)*a        

→       a = (F₁ - F₂) / (M₁  +M₂ ) = (12150-12285)N/(9*68+9*73)Kg

→       a = - 0.106 m/s^2            (←)

(b) What is the tension in the section of rope between the teams?

If we apply  ∑Fx = M*a   for the team 1

F₁ - T = - M₁*a  ⇒   T = F₁ + M₁*a  

⇒   T = 12150 N + (9 * 68 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

If we choose the team 2 we get

- F₂ + T = - M₂*a  ⇒   T = F₂ - M₂*a  

⇒   T = 12285 N - (9 * 73 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

4 0
3 years ago
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