Force and Laws of Motion-Exercise
Physics - Exercise
Q1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on
Solution:Yes, an object can have a non-zero velocity with zero net external force.
Let external force \(F\Rightarrow\) By Newton's second law of Motion, F can be written as $$\begin{aligned}F=ma\\ 0=ma\\ \Rightarrow a=0\end{aligned}$$ \(\Rightarrow\) acceleration zero does not mean velocity is zero, in the absence of an External force, an object may move with a constant velocity (Law of Inertia).
Q2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Explanation:Due to inertia, when the carpet is beaten with a stick, it starts to move, but the dust particles on it tend to remain at rest. Because the dust does not immediately follow the motion of the carpet, it loses contact with the carpet and comes off its surface.
Q3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Explanation:It is advised to tie any luggage kept on the roof of a bus with a rope because if the bus starts moving suddenly or the driver applies the brakes, the bus will change its motion, but the luggage will tend to resist this change due to inertia, which may cause it to fall off the roof.
Q4. A batsman hits a cricket ball, which then rolls on a level ground. After covering a short
distance, the
ball comes to rest. The ball slows to a stop because
(a) The batsman did not hit the ball hard enough.
(b) Velocity is proportional to the force exerted on the ball.
(c) There is a force on the ball opposing the motion.
(d) There is no unbalanced force on the ball, so the ball would want to come to rest.
(c) There is a force on the ball opposing the motion.\(\quad\Rightarrow\quad\color{blue} Correct\)
On the ground, friction acts opposite to motion. Also, air resistance resists motion. These opposing forces gradually reduce the velocity to zero.
Q5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
Solution:Using First Equation of Motion \((u=0,~t=20s)\) $$\begin{aligned}v&=u+at\\v&=0+20a \end{aligned} $$ $$v=20a\tag{1}$$ Uing third equation of Motion \(v^2 - u^2 = 2as\) $$\begin{aligned}v^{2}-u^{2}&=2as \\ v^{2}-0^{2}&=2\cdot a\cdot 400\\ \end{aligned}$$ $$v^{2}=800a\tag{2}$$ putting value of \(v\) from equation (1) to equation (2) $$\begin{aligned}\left( 20a\right) ^{2}&=800a\\ 400a^{2}&=800\\ a&=\dfrac{800}{400}\\ a&=2 m/s^2\end{aligned}$$ Force on th truck of mass \((m)=7 ~tonnes\) can be given as $$\begin{aligned} F&=ma\\ &=7000\times 2\\ &=14000 N\end{aligned}$$
Q6. A stone of 1 kg is thrown with a velocity of 20 m s–1 across the frozen surface of a lake, and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Solution:$$\begin{aligned}m&=1kg\\ u&=20ms^{-1}\\ s&=50m\\ v&=0\end{aligned}$$ Using the third equation of Motion \(v^2-u^2=2as\) $$\begin{aligned} v^{2}&=u^{2}+2as\\ 0^{2}&=\left( 20\right) ^{2}+2\cdot a\cdot 50\\ \Rightarrow 100a&=-20\times 20\\ a&=-4\times m/s^{2} \end{aligned}$$ $$\color{blue}\text{(-ve sign shows negative acceleration }\Rightarrow \text{retardation)}$$ Frictional force between the stone and ice can be given as $$\begin{aligned} F&=ma\\ 4&=1\cdot (-4)\\ &=-4N\end{aligned}$$
Q7. An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If
the engine
exerts a force of 40000 N, and the track offers a friction force of 5000 N, then
calculate:
(a) the net accelerating force and
(b) The acceleration of the train.
\(\begin{aligned}\text{Wt of Engine} &= 8000kg\\ \text{Wt of one wagon} &= 2000kg\\ \text{therefore, wt of 5 Wagon} = 2000 \times 5 &= 10,000kg\\ \text{total wt of train = 10000 + 8000 }&=18000kg\\ \text{Force Exerted by Engine}&=40000N\\ \text{Friction Force}&=5000N\\ \text{Net Force in moving direction = 40000 - 5000} &=35000N\\ \text{Total Mass }M&=18000kg\\ \text{Net Accelerating Force}&=35000N \end{aligned}\\\) The acceleration of the train can be given as $$\begin{aligned} F&=ma\\ \Rightarrow a&=F/m\\ &=\dfrac{35000}{18000}\\ &=\dfrac{35}{18}\\ &\approx 1.944ms^{-2}\end{aligned}$$
Q8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of \(1.7m/s^2\) ?
Solution:Mass \((m)\) of automobile = 1500kg $$\begin{aligned} a &= 1.7m/s^2 \text{ -(given)}\end{aligned}$$ The force between the vehicle and the road may be given as $$\begin{aligned} F&= ma\\ &= 1500 \times 1.7\\ &= 2550.0\\ &= 2550N\end{aligned}$$
Q9. What is the momentum of an object of mass \(m\), moving with
a velocity \(v\)?
(a) \((mv)^2\\\)
(b) \(mv^2\\\)
(c) \(\frac{1}{2} mv^2\\\)
(d) \(mv\)
(d) \(p=mv\Rightarrow\) is Correct
Q10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Solution:The friction force exerted on an object moving at constant velocity is equal in magnitude and opposite in direction to the applied force.
Since the cabinet is moving at constant velocity under a horizontal force of 200 N, the friction force must be 200 N, opposing the motion.
Q11. According to the third law of motion, when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Explanation:The student’s logic that “the two opposite and equal forces cancel each other” is incorrect because the two forces referred to in Newton’s third law act on different objects, not on the same object. According to Newton’s third law, if you push on the truck, the truck pushes back on you with an equal and opposite force. These forces are action-reaction pairs acting on two separate bodies— the push is on the truck, and the reaction is on the person pushing.
The reason the truck does not move is not because these two forces cancel out on the truck itself, but because the net force on the truck is zero. The truck is at rest because the forces acting on it (like friction between the truck’s wheels and the ground, or the truck’s own weight supported by the ground) balance the applied force, resulting in no acceleration.
Q12. A hockey ball of mass 200 g travelling at 10 \(m s^{-1}\) is struck by a hockey stick so as to return it along its original path with a velocity at 5 \(m s^{–1}\). Calculate the magnitude of the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Solution: Given that:Mass of the hockey ball \(m=200g\\\) Initial Velocity \(u=10m/s\\\) Final Velocity \(v=-5m/s\) as ball will move in opposite direction
Change in momentum \(\Delta p = m\Delta v\) $$\begin{aligned} \triangle p&=m\Delta v\\ &=\dfrac{200}{1000}\left( u-\left( -v\right) \right) & \color{blue}\left(\because\quad 200 g = \frac{200}{1000} kg\right) \\ &=\dfrac{200}{1000}\left( 10-\left( -5\right) \right) \\ &=\dfrac{200}{1000}\left( 10+5\right) \\ &=\dfrac{200\times 15}{1000}\\ &=\dfrac{200\times 15}{1000}\\ &=\dfrac{3000}{1000}\\ &=3N\end{aligned}$$
Q13. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s–1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Solution:Mass of bullet \((m)=10 g\\\) Initial Velocity \(u = 150m/s\\\) Time when bullet comes to rest \(t = 0.03s\\\) Final Velocity of bullet \(v = 0\\\) Acceleration of the bullet can be calculated by using First Equation of Motion $$\begin{aligned} v&=u+at\\ 0&=150+a\cdot 0.03\\ -0.03a&=150\\\\ a&=-\dfrac{150}{0\cdot 03}&\color{blue}\text{(-ve value shows retardation)}\\\\ a&=-\dfrac{150\times 100}{3}\\\\ =&-5000m/s\end{aligned}$$ Force Exerted on the wooden block $$\begin{aligned} F&=ma\\ &=10\times 10^{-3}\times 5\times 10^{3}\\ &=50N\end{aligned}$$ Let Distance penetrated by the bullet be \(s\), therefore, using third equation of motion we ca find the distance $$\begin{aligned} v^{2}-u^{2}&=2as\\\\ 0-\left( 150\right) ^{2}&=2\times (-5\times 10^{3})\cdot s\\\\ \Rightarrow s&=\dfrac{150\times 150}{2\times 5\times 1000}\\\\ s&=2.25m\end{aligned}$$
Q14. An object of mass 1 kg travelling in a straight line with a velocity of \(10 m s^–1\) collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Solution:Mass \((m)\) of the object = 1kg
Initial velocity \(u= 10m/s\)
Mass of wood \(M = 5kg\)
Total mass $$\begin{aligned}&= m + M\\ &= 1kg + 5kg\\ &= 6kg\end{aligned}$$ Moment just before the Impact $$\begin{aligned}p &= m u\\ &= 1\times 10\\ &= 10 kg m/s\end{aligned}$$ By conservation of momentum $$\begin{aligned} p_{1}&=p_{2}\\ m_1u&=m_2v&\left(m_1=m\Rightarrow1kg,~m_2=m+M\Rightarrow 6kg\right)\\ 1\cdot 10&=6.v\\\\ v_{2}&=\dfrac{10}{6}\\\\&=\dfrac{5}{3}\\\\ &\approx1\cdot 66m/s \end{aligned}$$ Moment just after the Impact $$\begin{aligned}p &= (m+M) v\\\\ &= 6\times \frac{5}{3}\\\\ &= 10 kg m/s\end{aligned}$$ By conservation of momentum
Q15. An object of mass 100 kg is accelerated uniformly from a velocity of \(5 m s^–1\) to \(8 m s^–1 \)in \(6 s\). Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object
Solution:Mass of the object \((m) = 100 kg\\\) Initial velocity \(u = 5m/s\\\) Final velocity \(v = 8 m/s\\\) Time taken to accelerate \(t = 6s\\\) By first equation of motion $$\begin{aligned}v&=u+at\\ 8&=5+a\cdot 6\\ 8-5&=6a\\ \Rightarrow 6a&=3\\ a&=3/6\\ =&0.5ms^{-2}\end{aligned}$$ Initial momentum $$\begin{aligned}p &= mu\\ &= 100\times 5\\ &= 500 Kgm/s\end{aligned}$$ Final Momentum $$\begin{aligned}p &= mv\\ &= 100\times 8\\ &= 800 Kgm/s\end{aligned}$$ Force Exerted on the object: $$\begin{aligned} F&=ma\\&= 100\times 0.5\\&=50N\end{aligned}$$
Q16. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Comments:Kiran’s suggestion that the insect suffered a greater change in momentum than the motorcar is incorrect because the change in momentum of the insect and the motorcar must be equal in magnitude due to the conservation of momentum, though the insect’s velocity changes much more since it has a much smaller mass than the motorcar.
Akhtar’s suggestion that the motorcar exerts a larger force on the insect because of its larger velocity is also wrong. By Newton’s third law, the forces on the insect and motorcar during the collision are equal in magnitude and opposite in direction. The insect dies not because of a larger force on it but because its small mass and sudden velocity change cause a large acceleration and damage.
Rahul’s explanation that both the motorcar and the insect experience the same force and the same magnitude of change in momentum is correct. However, the directions of these momentum changes are opposite. The motorcar’s large mass means a very small velocity change, while the insect’s small mass means a larger velocity change. mass leads to a large velocity change upon impact, causing fatal injury to the insect.
How much momentum will a dumbbell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be \(10 m s^–2\)
Solution:Given that:
Initial Velocity \(u=0\\\) Downward aceleration \(=10m/s^2\\\) Using third equation of Motion $$\begin{aligned}v^{2}-u^{2}&=2as\\ v^{2}-0^{2}&=2\cdot 10\cdot 0.8\\ v^{2}&=2\times 8\\ v&=\sqrt{16}\\ v&=4m/s\end{aligned}$$ therefore, Momentum of the dumbell $$\begin{aligned}p&=mv\\&=10\times 4\\&=40kgm/s\end{aligned}$$
