Physics

Which of the following statements give characteristics of contact forces ?

• 1. It appears between an object when it is in contact with some other object
• 2. It satisfies the third law of motion
• 3. It may appear between a pair of solid and fluid

Select the answer using the code given below :

[amp_mcq option1=”1 and 3 only” option2=”2 and 3 only” option3=”1 and 2 only” option4=”1, 2 and 3″ correct=”option4″]

The correct answer is D, as all three statements are correct characteristics of contact forces.

– Contact forces are forces that arise when two objects are in physical contact with each other. Examples include normal force, friction, tension, and air resistance. Statement 1 accurately describes this.
– Newton’s Third Law of Motion states that for every action, there is an equal and opposite reaction. All forces in nature, including contact forces, obey this law. For instance, if object A exerts a contact force on object B, then object B simultaneously exerts an equal and opposite contact force on object A. Statement 2 is correct.
– Contact forces can exist between solids, between a solid and a fluid (liquid or gas), or between different fluids. For example, buoyancy and drag force are contact forces between a solid and a fluid, or within a fluid. Statement 3 is correct.

Forces are broadly classified into contact forces and non-contact forces (like gravitational, electromagnetic, and nuclear forces) which act over a distance. The statements correctly describe the nature and properties of contact forces.

An astronaut whose weight on the Earth is 600 N experiences weightlessness on International Space Station orbiting around the Earth. It means that

[amp_mcq option1=”acceleration of the astronaut is zero” option2=”normal reaction of the space station floor on the astronaut is zero” option3=”gravitational pull of earth on the astronaut is zero” option4=”space station applies a centrifugal force on the astronaut” correct=”option2″]

The correct answer is B. An astronaut experiences weightlessness on the International Space Station (ISS) because the normal reaction force from the space station floor on the astronaut is zero.

– Weight is the force exerted on an object due to gravity. On Earth, we feel our weight because of the normal force exerted by the surface supporting us, which balances gravity.
– The ISS and everything in it, including the astronaut, are constantly in freefall around the Earth. They are orbiting because they have a high tangential velocity while simultaneously accelerating towards the Earth due to gravity.
– In freefall, there is no supporting surface providing a normal reaction force to counteract gravity. The feeling of weight comes from this reaction force. Since this support force is absent, the astronaut feels weightless.
– The gravitational pull of Earth on the astronaut is not zero in orbit; it is still significant (around 90% of Earth’s surface gravity at ISS altitude) and is what keeps the ISS in orbit.
– The astronaut is accelerating towards the Earth (centripetal acceleration required for circular motion), so their acceleration is not zero.
– Centrifugal force is a fictitious force experienced in a rotating frame of reference; it’s not a real force causing weightlessness.

The state of apparent weightlessness in orbit is often referred to as microgravity. It is not due to the absence of gravity, but rather the state of continuous freefall. The ISS is continuously falling towards Earth, but its high orbital speed causes it to miss the Earth, resulting in an orbit.

Shown in the figure are two hollow cubes C₁ and C₂ of negligible mass partially filled (depicted by darkened area) with liquids of densities ρ₁ and ρ₂, respectively, floating in water (density ρw). The relationship between ρ₁, ρ₂ and ρw is

[amp_mcq option1=”ρ₂ < ρw < ρ₁" option2="ρ₂ < ρ₁ < ρw" option3="ρ₁ < ρ₂ < ρw" option4="ρ₁ < ρw < ρ₂" correct="option4"]

The correct answer is D) ρ₁ < ρw < ρ₂.

Both cubes are floating in water. According to Archimedes’ principle, a floating object displaces a volume of fluid whose weight is equal to the weight of the object. The weight of each hollow cube is solely due to the liquid inside, as the cube itself has negligible mass. Let A be the base area and H be the total height of the cubes (assuming they are identical). Let h₁ and h₂ be the submerged heights in water, and H₁_liquid and H₂_liquid be the heights of the liquid inside.
For Cube C₁:
Weight of liquid inside = (Volume of liquid inside) * ρ₁ * g = (A * H₁_liquid) * ρ₁ * g.
Buoyant force = (Volume submerged) * ρw * g = (A * h₁) * ρw * g.
Since it’s floating, (A * H₁_liquid) * ρ₁ * g = (A * h₁) * ρw * g, which simplifies to H₁_liquid * ρ₁ = h₁ * ρw, or ρ₁ = (h₁ / H₁_liquid) * ρw.
From the figure, the submerged height h₁ is significantly less than the height of the liquid inside H₁_liquid. Therefore, (h₁ / H₁_liquid) < 1, which implies ρ₁ < ρw. For Cube C₂: Weight of liquid inside = (A * H₂_liquid) * ρ₂ * g. Buoyant force = (A * h₂) * ρw * g. Since it's floating, (A * H₂_liquid) * ρ₂ * g = (A * h₂) * ρw * g, which simplifies to H₂_liquid * ρ₂ = h₂ * ρw, or ρ₂ = (h₂ / H₂_liquid) * ρw. From the figure, the submerged height h₂ is significantly greater than the height of the liquid inside H₂_liquid. Therefore, (h₂ / H₂_liquid) > 1, which implies ρ₂ > ρw.

Combining the results, we have ρ₁ < ρw and ρ₂ > ρw. This means ρ₁ is less than the density of water, while ρ₂ is greater than the density of water. Therefore, the relationship between the densities is ρ₁ < ρw < ρ₂.

A 100 g sphere is moving at a speed of 20 m/s and collides with another sphere of mass 50 g. If the second sphere was at rest prior to the collision and the first sphere comes to rest immediately after the collision, considering the collision to be elastic, the speed of the second sphere would be

[amp_mcq option1=”10 m/s” option2=”20 m/s” option3=”30 m/s” option4=”40 m/s” correct=”option4″]

Given:
Mass of first sphere (m1) = 100 g = 0.1 kg
Initial velocity of first sphere (u1) = 20 m/s
Mass of second sphere (m2) = 50 g = 0.05 kg
Initial velocity of second sphere (u2) = 0 m/s (at rest)
Final velocity of first sphere (v1) = 0 m/s (comes to rest)
Final velocity of second sphere (v2) = ?

For any collision in an isolated system, momentum is conserved.
Conservation of Momentum: m1*u1 + m2*u2 = m1*v1 + m2*v2
(0.1 kg) * (20 m/s) + (0.05 kg) * (0 m/s) = (0.1 kg) * (0 m/s) + (0.05 kg) * v2
2 + 0 = 0 + 0.05 * v2
2 = 0.05 * v2
v2 = 2 / 0.05 = 2 / (5/100) = 2 * (100/5) = 2 * 20 = 40 m/s.

While the question states the collision is elastic, the conditions provided (m1=100g, m2=50g, u1=20m/s, u2=0, v1=0) are mathematically inconsistent with a perfectly elastic collision. For an elastic collision with u2=0, v1 = ((m1 – m2) / (m1 + m2)) * u1. Substituting the values, v1 = ((0.1 – 0.05) / (0.1 + 0.05)) * 20 = (0.05/0.15) * 20 = (1/3) * 20 = 20/3 m/s, which is not 0. This means that either the collision was not elastic, or the first sphere did not come to rest.
However, given the multiple-choice format and the need to derive one of the options, the value v2 = 40 m/s is obtained directly from the conservation of momentum equation using the provided initial and final states. It is common in flawed physics problems in exams that one must rely on the most directly applicable principle (momentum conservation) and the given numbers, even if they contradict another stated condition (elasticity). Therefore, assuming the initial conditions and the final velocity of the first sphere are as stated, and momentum is conserved, the speed of the second sphere is 40 m/s.

– In any collision in an isolated system, momentum is conserved.
– For an elastic collision, kinetic energy is also conserved.
– The given conditions in this question are contradictory for a perfectly elastic collision.
– The answer is derived using conservation of momentum and the provided values, including the final velocity of the first sphere.

If the collision were truly elastic with the given masses and initial velocities, the final velocities would be v1 = 20/3 m/s and v2 = 80/3 m/s. Since 40 m/s is an option and is derivable from the given conditions via momentum conservation, it is the most likely intended answer despite the inconsistency.

Which one of the following graphs represents the equation of motion v = u + at; where all quantities are non-zero and symbols carry their usual meanings ?

[amp_mcq option1=”(a)” option2=”(b)” option3=”(c)” option4=”(d)” correct=”option3″]

The equation of motion is given by v = u + at, where:
– v is the final velocity
– u is the initial velocity
– a is the acceleration (constant)
– t is the time
This equation is in the form of a linear equation y = mx + c, where v is the dependent variable (y), t is the independent variable (x), ‘a’ is the slope (m), and ‘u’ is the y-intercept (c). The problem states that all quantities (u, a, v, t) are non-zero, except t=0 at the start.
A graph of v versus t for this equation should be a straight line.
Since ‘u’ is non-zero, the line will intersect the v-axis (at t=0) at a point other than the origin.
Since ‘a’ is non-zero, the slope of the line will not be zero (the line is not horizontal).
Considering these points, graph (c) represents a straight line with a positive y-intercept (u > 0) and a positive slope (a > 0). While ‘a’ could be negative (negative slope) or ‘u’ could be negative (negative intercept), option (c) is the only graph that shows a straight line with a non-zero intercept and a non-zero slope, fitting the general form v = u + at with non-zero u and a. Graphs (a) and (d) are not straight lines, and graph (b) is a straight line passing through the origin, which would represent v = at (i.e., u=0), contrary to the condition that u is non-zero.

– The equation v = u + at represents linear motion with constant acceleration.
– A graph of velocity (v) versus time (t) for this equation is a straight line.
– The y-intercept of the line is the initial velocity (u).
– The slope of the line is the acceleration (a).

If ‘a’ were zero, the graph would be a horizontal line at v = u. If ‘u’ were zero, the line would pass through the origin (v = at). The shape of the line (upward or downward slope) depends on the sign of acceleration ‘a’. Option (c) shows a positive slope (acceleration).