Question

The velocity of a 140 gram ball changes from < 7, −7, 0 > m/s to < 7.3, −7.4, 0 > m/s in 0.07 s, due to the gravitational attraction of the Earth and to air resistance. (a) What is the acceleration of the ball?

Answer 1

Complete question:

The velocity of a 140 gram ball changes from < 7, −7, 0 > m/s to < 7.3, −7.4, 0 > m/s in 0.07 s, due to the gravitational attraction of the Earth and to air resistance.

(a) What is the acceleration of the ball?

(b) What is the rate of change of momentum of the ball?

(c) What is the net force acting on the ball?

Answer:

Part (i) the acceleration of the ball is 10m/s²

Part(ii) the rate of change of momentum of the ball is 1.4N

Part (iii) the net force acting on the ball is 0.028 N

Explanation:

Given;

mass of the ball = 140 g = 0.14 kg

change in initial velocity of the ball, u = (7.3 - 7) m/s = 0.3 m/s

change in final velocity of the ball, v =  ( -7.4 + 7) m/s = -0.4 m/s

time of motion = 0.07s

Part (i) the acceleration of the ball

a = Δv/t

$a = \frac{v-u}{t} \\\\a = \frac{-0.4-0.3}{0.07}\\\\a =\frac{-0.7}{0.07} = -10\frac{m}{s^2}\\\\|a| = 10\frac{m}{s^2}$

Part(ii) the rate of change of momentum of the ball

From newton's second law of motion; "rate of change of momentum is directly proportional to the applied force.

$F =\frac{m(V-u)}{t} = ma\\\\F = 0.14 X 10= 1.4N$

Part (iii) the net force acting on the ball

$F_{Net} = F_a - F_g\\\\F_{Net} = M_a -M_g\\\\ F_{Net} = M(a -g)\\\\F_{Net} = 0.14(10-9.8)\\\\F_{Net} = 0.028 N$

Answer 2

The acceleration of the ball is 7 m/s².

Acceleration:

The initial velocity of ball is <7,-7,0> m/s and the final velocity is <7.3,-7.4,0> m/s.

Change in velocity Δv =  <7.3,-7.4,0> -  <7,-7,0> = <0.3,-0.4,0>

The magnitude of change in velocity is :

$|\Delta v|=\sqrt{(0.3)^2+(0.4)^2}\\\\ |\Delta v|=0.5\;m/s$

Acceleration is defined as the rate of change of velocity, that is:

$a=\frac{|\Delta v|}{\Delta t}\\\\a=\frac{0.5}{0.07}\;m/s^2\\\\a\approx7\;m/s^2$

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