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Chapter 8: Problem 2

A stone weighing \(S \mathbb{N}\) is thrown vertically upwards, with velocity \(80 \mathrm{~ms}^{-} .\) What is its velocity after two seconds and after twenty seconds? \(\left(\right.\) take \(\left.g=10 \mathrm{~ms}^{-2}\right)\)

Short Answer

Expert verified
After 2 seconds, the velocity is 60 m/s upwards. After 20 seconds, the velocity is 120 m/s downwards.

Step by step solution

01

- Understand the Problem

A stone is thrown vertically upwards with an initial velocity of 80 m/s. The acceleration due to gravity is given as 10 m/s虏. The task is to find the velocity after 2 seconds and after 20 seconds.
02

- Apply the Kinematic Equation

Use the equation for velocity under constant acceleration: \[ v = u + at \]where \( v \) is the final velocity, \( u \) is the initial velocity (80 m/s, since it is thrown upwards), \( a \) is the acceleration (which is -10 m/s虏 due to gravity acting downwards), and \( t \) is time.
03

- Calculate Velocity After 2 Seconds

Substitute the values into the equation for 2 seconds:\[ v = 80 + (-10) \times 2 \]\[ v = 80 - 20 \]\[ v = 60 \text{ m/s} \]Therefore, the velocity after 2 seconds is 60 m/s upwards.
04

- Calculate Velocity After 20 Seconds

Substitute the values into the equation for 20 seconds:\[ v = 80 + (-10) \times 20 \]\[ v = 80 - 200 \]\[ v = -120 \text{ m/s} \]Therefore, the velocity after 20 seconds is -120 m/s, which means 120 m/s downwards.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

kinematic equations
Kinematic equations are essential in solving problems related to motion. They help us understand how different parameters like velocity, acceleration, time, and displacement are related.
Kinematic equations assume constant acceleration and allow us to predict future motion states based on initial conditions. In this exercise, we used the kinematic equation \(v = u + at\) to find the final velocity.
This equation directly links the initial velocity, the acceleration, and the time elapsed to the final velocity. Remember, when solving any motion problem, always identify which kinematic equation fits the data and unknowns you have.
initial velocity
Initial velocity is the velocity of an object at the start of the time interval. In our exercise, the stone is thrown vertically upward with an initial velocity (u) of 80 m/s.
Understanding initial velocity is crucial because it sets the starting point for any changes to the object's velocity. It鈥檚 important to annotate whether the initial velocity is positive (upwards or right) or negative (downwards or left), as it affects subsequent calculations.
In problems involving vertical motion, the initial velocity is often provided or can be calculated. Always make sure to correctly incorporate its value into the kinematic equations.
acceleration due to gravity
Acceleration due to gravity (g) is a constant value of 9.8 m/s虏, but for simplicity, we often round it to 10 m/s虏 in problems. It represents the rate at which an object accelerates downwards due to Earth's gravitational pull.
In the exercise, gravity acts as a decelerating force when the stone is thrown upwards, hence it鈥檚 taken as -10 m/s虏. This negative acceleration indicates a reduction in velocity as the stone ascends.
Always consider the direction of acceleration - it's downward, so in problems involving upward motion, it's negative. This is critical when plugging values into kinematic equations.
final velocity calculation
Final velocity (\textit v) is the velocity of an object at the end of the time interval. It can be computed through kinematic equations when initial velocity, time, and acceleration are known.
In our exercise, to find the final velocity after 2 and 20 seconds, we used the equation \(v = u + at\). By substituting u = 80 m/s, a = -10 m/s虏, and t = 2 s, we found that the velocity after 2 seconds is 60 m/s upward. Similarly, for t = 20 s, the final velocity is -120 m/s, indicating downward motion.
Final velocity helps to predict where and how fast the object will be moving after a certain period, pivotal in many physics problems.

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Most popular questions from this chapter

28) The masses of three perfectly elastic spheres \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are \(M, M\) and \(m\) respectively \((M>m)\). The spheres are initially at rest with their centres in a straight line, \(\mathrm{C}\) lying between \(\mathrm{A}\) and \(\mathrm{B}\). If \(\mathrm{C}\) is given a velocity towards \(\mathrm{A}\) along the line of centres, show that after colliding first with \(\mathrm{A}\) and then with \(\mathrm{B}\) it will not collide a second time with \(\mathrm{A}\) if \(M<(\sqrt{5}+2) m\). Find the ratios of the kinetic energies of the three spheres after the second collision and verify that no energy has been lost.

A smooth sphere \(\mathrm{A}\) of mass \(2 m\), moving on a horizontal plane with speed \(u\), collides directly with another smooth sphere B of equal radius and of mass \(m\), which is at rest. If the coefficient of restitution between the spheres is \(e\), find their speeds after impact. The sphere B later rebounds from a perfectly elastic vertical wall, and then collides directly with A. Prove that after this collision the speed of B is \(9(1+e)^{2} u\) and find the speed of \(\mathrm{A}\).

Two particles \(\mathrm{A}\) and \(\mathrm{B}\) collide directily head-on and bounce. Find their speeds immediately after impact. (a) The mass of \(\mathrm{A}\) is twice the mass of B. (b) Just before impact the speed of \(\mathrm{A}\) is \(4 \mathrm{~ms}^{-1}\) and that of \(\mathrm{B}\) is \(3 \mathrm{~ms}^{-1} .\) (c) No kinetic energy is lost by the impact.

A particle of mass \(m\) is thrown vertically upwards with speed \(u\) from a point A on the ground. Simultaneously an identical particle is thrown vertically downwards also with speed \(u\) from a point B vertically above \(\mathrm{A}\) and at a height \(h\) above the ground \(\left(h<4 u^{2} / g\right)\). On impact the particles adhere and move subsequently as a single particle. Calculate the loss in kinetic energy caused by the impact and the speed of the combined particle on reaching the ground.

A particle of mass \(4 \mathrm{~kg}\) is attached to one end \(\mathrm{X}\) of a light inextensible string which passes over a smooth light pulley and supports particies of masses \(2 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) at the other end \(\mathrm{Y}\). The end \(\mathrm{X}\) is held in contact with a horizontal table at a depth \(6 \mathrm{~m}\) below the pulley, both portions of the string being vertical and the particles at \(\mathrm{Y}\) hanging freely. The system is released from rest. When \(\mathrm{Y}\) has descended a distance of \(2.5 \mathrm{~m}\), the particle of mass \(2 \mathrm{~kg}\) is disconnected and begins to fall freely. Calculate the greatest height reached by \(\mathrm{X}\) above the table and the momentum of the \(4 \mathrm{~kg}\) particle when it strikes the table. Take \(g\) to be \(9.81 \mathrm{~m} / \mathrm{s}^{2}\)
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