Question
Question: A player stops a football weighing \[0.5\,kg\] which comes flying towards him with a velocity of \(1...
A player stops a football weighing 0.5kg which comes flying towards him with a velocity of 10msec−1 .If the impact lasts for 501th second and the ball bounces back with a velocity of 15msec−1 then the average force involved is:
A. 250N
B. 1250N
C. 500N
D. 625N
Solution
In order to solve this question, we will use the concept of Newton’s second law of motion which states that force applied on a body is always equals to the rate of change of momentum of the body, where momentum is defined as the product of mass and velocity of the body.
Formula used:
Momentum of body is calculated as,
p=mv
where m is mass and v is the velocity of the body.
Force due the Newton’s second law of motion is written as,
∣F∣=dt∣dp∣
Complete step by step answer:
According to the parameters given in the question, we have mass of the football as m=0.5kg. Let Initial velocity of the ball is denoted by u=10msec−1. Now, after bouncing back the final velocity of the ball will be in opposite direction to that of initial velocity direction hence, final velocity will be taken as negative
v=−15msec−1
Let initial momentum denoted as pinitial=mu so we have,
pinitial=0.5×10
⇒pinitial=5kgmsec−1→(i)
Let final momentum denoted as pfinal=mv so we have,
pfinal=−0.5×15
⇒pfinal=−7.5kgmsec−1→(ii)
Now, let change in momentum is dp=pfinal−pinitial
From equations (i)and(ii) we have,
dp=−7.5−5
⇒dp=−12.5kgmsec−1
Since the magnitude of impact is given as dt=501sec
Now, using the formula of force as ∣F∣=dt∣dp∣ we have,
∣F∣=12.5×50
∴∣F∣=625N
Hence, the correct option is D.
Note: It should be remembered that, while solving such questions keep the direction of velocity in mind, every time a body reflects or bounce back the direction of velocity gets reversed and also Force and momentum both are a vector quantity and in simpler terms force on a body is just the product of mass and acceleration of the body written as F=ma.