Physics, 22.01.2021 14:00 1232444553

A 0.061 kg golf ball is struck by a golf club with a force of 299.4 N. The change in velocity of the ball is 54.3 m/s. How much time is the ball in contact with the club?

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Answer from: 20011854

The time the ball is in contact with the club is approximately 0.011063 seconds

Explanation:

The question is with regards to Newton's second law of motion which states that a force is equal to the rate of change of momentum produced

The given parameters are;

The mass of the ball, m = 0.061 kg

The force with which the ball is struck with the golf club, F = 299.4 N

The change in velocity of the ball, Δv = 54.3 m/s

By Newton's second law, we have;

$F_{net} = \dfrac{\Delta P}{\Delta t} = \dfrac{m \times \Delta v}{\Delta t}$

Where;

Δt = The time it takes the momentum of the object to change = The time the ball is in contact with the club

Substituting the known values, we get;

$F_{net} = F =299.4 = \dfrac{0.061 \times 54.3}{\Delta t}$

$\therefore \Delta t = \dfrac{0.061 \times 54.3}{299.4} \approx 0.011063$

The time the ball is in contact with the club = Δt ≈ 0.011063 seconds.

Answer from: Quest

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s