a ball of mass 10g is initially moving with velocity of 50m/s .on applying a constant force on the ball for 2s, it acquires a velocity of 70m/s .calculate. initial momentum of the ball.final momentum of the ball.rate of change of momentum. acceleration of the ball.magnitude of the force applied.
Answers
m = 10 kg u = 50 m/s v = 70 m/s t = 2s a = ? F = ? (a) Pi = m•u . = (10•50)kg m/s . = 500 kg m/s (b) Pf = m•v . = (10•70) kg m/s . = 700 kg m/s (c) a = (v-u)/t . = 10 m/s² (d) F = m•a . = 50kg x 10m/s² . = 500 kg m/s² . = 500N .
Concept:
Newton's second law of motion states that Force can be defined as the rate of change of momentum.
Given:
Mass of the ball, m = 10g, m= 0.01 kg
Initial velocity, u =50 m/s
Force applied for time, t = 2 s
Final velocity, v = 70 m/s
Find:
The initial momentum of the ball, the rate of change of the momentum, acceleration, and the magnitude of force applied by the ball.
Solution:
Momentum can be written as the product of mass and velocity.
Initial momentum = mass × Initial velocity
P = m × u = 0.01 × 50 = 0.5 kg m /s
The rate of change of momentum can be defined as the force applied which is equal to the product of mass and acceleration,
Change in velocity per unit time is known as Acceleration.
a = (v-u) / t = (70-50)/2 = 20/2 = 10 ms⁻²
So, Rate of change of momentum,
ΔP = m × a = 0.01 × 10 = 0.1 kg m s⁻²
As calcuated,
a = (v-u) / t = (70-50)/2 = 20/2 = 10 ms⁻²
As the magnitude of force applied is equal to the rate of change of momentum,
F = ΔP = 0.1 kg m s⁻²
Hence, the initial momentum of the ball is 0.5 kg m /s, the rate of change of the momentum is 0.1 kg m s⁻², the acceleration of the ball is 10 ms⁻², and the magnitude of force applied is 0.1 kg m s⁻².
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