Kinematics and Kinetics MCQ Quiz - Objective Question with Answer for Kinematics and Kinetics - Download Free PDF

Concept:

We analyze the motion of a particle with variable acceleration to determine its velocity when the acceleration becomes zero.

Given:

• Acceleration function: \( a = 10 - x \)
• Initial conditions: at \( t = 0 , ~ x = 0 ,~ and~ velocity ~v = 0 \)

Step 1: Find the position when acceleration is zero

Set \( a = 0\)

\( 10 - x = 0 \)

\( x = 10 \, \text{m} \)

Step 2: Relate acceleration to velocity and position

Using the chain rule, acceleration can be expressed as:

\( a = \frac{dv}{dt} = v \frac{dv}{dx} = 10 - x \)

Separate variables and integrate:

\( v \, dv = (10 - x) \, dx \)

Step 3: Integrate both sides

\( \int_{0}^{v} v \, dv = \int_{0}^{10} (10 - x) \, dx \)

\( \frac{v^2}{2} \bigg|_{0}^{v} = 10x - \frac{x^2}{2} \bigg|_{0}^{10} \)

\( \frac{v^2}{2} = 100 - 50 = 50 \)

\( v^2 = 100 \)

\( v = 10 \, \text{m/s} \)

Concept:

When a body moves along a curved path with constant tangential acceleration, the total acceleration has two components:

1. Tangential acceleration due to change in speed.

2. Normal (centripetal) acceleration due to change in direction, given by:

\( a_n = \frac{v^2}{r} \)

Calculation: 

Given:

Final speed after 60 seconds: v = 18 km/h = 5 m/s

Radius of curve: r = 200 m

Time = 60 s, Initial speed = 0 (starts from rest)

Step 1: Calculate tangential acceleration

\( a_t = \frac{v - u}{t} = \frac{5 - 0}{60} = \frac{1}{12}~m/s^2 \)

Step 2: Speed at 30 seconds:

\( v = a_t \times t = \frac{1}{12} \times 30 = 2.5~m/s \)

Step 3: Normal acceleration

\( a_n = \frac{v^2}{r} = \frac{(2.5)^2}{200} = \frac{6.25}{200} = 0.03125~m/s^2 \)

Concept:

To find the time of meeting of the two shots, we consider the relative motion of the projectiles in the horizontal direction.

Given:

Distance between guns, d = 40 m

Velocity of shot from Gun A, v_A = 350 m/s

Velocity of shot from Gun B, v_B = 300 m/s

Angle of projection, \(\theta = 30^\circ\)

Calculation:

Resolving velocities into horizontal components:

\( v_{Ax} = v_A \cos 30^\circ = 350 \times \frac{\sqrt{3}}{2} = 175\sqrt{3} \) m/s

\( v_{Bx} = v_B \cos 30^\circ = 300 \times \frac{\sqrt{3}}{2} = 150\sqrt{3} \) m/s

The relative velocity in the horizontal direction is:

\( v_{\text{relative}, x} = v_{Ax} + v_{Bx} = 175\sqrt{3} + 150\sqrt{3} = 325\sqrt{3} \) m/s

Using the formula,

\( \text{Time} = \frac{\text{Distance}}{\text{Relative Velocity}} \)

\( t = \frac{40}{325\sqrt{3}} \)

Approximating \(\sqrt{3} \approx 1.732\) ,

\( t = \frac{40}{325 \times 1.732} = \frac{40}{562.9} \approx \frac{4}{65}\) seconds

Concept:

When two bodies collide and move together after the collision, the principle of conservation of momentum is applied.

The equation for conservation of momentum is given by:

\( m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \)

Where:

• \( m_1 = 3 \) kg (mass of first sphere)
• \( v_1 = 5 \) m/s (velocity of first sphere before collision)
• \( m_2 = 2 \) kg (mass of second sphere)
• \( v_2 = 0 \) m/s (velocity of second sphere before collision, since it is at rest)
• \( v_f \) is the common velocity after collision

Calculation:

Applying the conservation of momentum equation:

\( (3 \times 5) + (2 \times 0) = (3 + 2) v_f \)

\( 15 = 5 v_f \)

\( v_f = \frac{15}{5} = 3 \) m/s

Explanation:

Step 1: Calculate the initial momentum of the bullet

The initial momentum of the bullet can be calculated using the formula:

Momentum (p) = Mass (m) × Velocity (v)

The mass of the bullet (m_bullet) is:

m_bullet = W_bullet / g = 0.5 N / 10 m/s² = 0.05 kg

Given the velocity of the bullet (v_bullet) = 810 m/s, the initial momentum of the bullet is:

p_bullet = m_bullet × v_bullet = 0.05 kg × 810 m/s = 40.5 kg·m/s

Step 2: Calculate the combined mass of the block and bullet system

• Weight of the block (W_block) = 40 N

The mass of the block (m_block) is:

m_block = W_block / g = 40 N / 10 m/s² = 4 kg

The combined mass of the block and bullet system (m_total) is:

m_total = m_block + m_bullet = 4 kg + 0.05 kg = 4.05 kg

Step 3: Calculate the velocity of the block and bullet system after the collision

Using the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

Total initial momentum = Total final momentum

p_bullet = m_total × v_final

Solving for v_final (the final velocity of the block and bullet system):

v_final = p_bullet / m_total = 40.5 kg·m/s / 4.05 kg = 10 m/s

When the block is moving on the rough horizontal plane, it is subjected to frictional force. The frictional force (f_friction) can be calculated using the formula:

f_friction = μ × N

where μ is the coefficient of friction and N is the normal force, Coefficient of friction (μ) = 0.3,

Normal force (N) = W_block + W_bullet = 40 N + 0.5 N = 40.5 N

The frictional force is:

f_friction = 0.3 × 40.5 N = 12.15 N

Using Newton's second law, we can calculate the deceleration (a) due to friction:

a = f_friction / m_total = 12.15 N / 4.05 kg = 3 m/s²

The block will decelerate to a stop due to this frictional force. We can use the kinematic equation to find the distance (d) travelled by the block:

v_final² = v_initial² - 2ad

Here, v_final = 0 (since the block comes to rest), v_initial = 10 m/s, and a = 3 m/s².

0 = (10 m/s)² - 2 × 3 m/s² × d

Solving for d:

100 = 6d

d = 100 / 6 = 16.67 m

Thus, the distance travelled by the block from its initial position is 16.67 meters.

Concept:

• Average velocity = total displacement/ total time duration
• Equation of motion:
• v = u + at
• v² = u² + 2as
• s = ut + 1/2 at2

Calculation:

Given:

Time interval = 5 s & 1 s, Initial velocity u = 0, and, Acceleration a = 2 m/s²

​When an object starts from rest, then the total distance covered in time 1 sec and 5 sec is,

s = ut + 1/2 at2

Object is at rest, so, u = 0 m/s.

\(s_2 - s_1 = \frac12 a(t_2^2-t_1^2)\)

\(s_2 - s_1 = \frac12 \times 2(5^2-1^2)\)

s₂ - s₁ = 24 m

Total time taken, t = t₂ - t₁ = 5 - 1 = 4 sec

Average velocity = total displacement/ total time duration

Average velocity = 24/4 = 6 m/s

Average velocity between time 1 s and 5 s = 6 m/s

Concept:

Rate of change of velocity is known as acceleration. Its unit is m/s2. It is a vector quantity.

a = change in velocity/time

Equations of motion:

• v = u + at
• v2 – u2 = 2as
• \(s = ut + \frac{1}{2}a{t^2}\)

Calculation:

Given:

u = 36 km/h = 10 m/s; S = 125 m ; v = 54 km/h = 15 m/s,  t = ?

v2 – u2 = 2as

\(a = \frac{{{v^2} - {u^2}}}{{2s}} = \frac{{{{15}^2} - {{10}^2}}}{{2 \times 125}} = 0.5\;m/s^2\)

v = u + at

\(t = \frac{{v - u}}{a} = \frac{{15 - 10}}{0.5} = 10\;sec\)

CONCEPT:

Centripetal Acceleration (a_c): 

• Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration.
• It always acts on the object along the radius towards the center of the circular path.
• The magnitude of centripetal acceleration,

\(a = \frac{{{v^2}}}{r}\)

Where v = velocity of the object and r = radius

Tangential acceleration (a_t):

• It acts along the tangent to the circular path in the plane of the circular path.
• Mathematically Tangential acceleration is written as

\(\overrightarrow {{a_t}} = \vec \alpha \times \vec r \)

Where α = angular acceleration and r = radius

CALCULATION:

Given – v = 10 m/s, r = 25 m and a_t = 3 m/s²

• Net acceleration is the resultant acceleration of centripetal acceleration and tangential acceleration i.e.,

\(a = \sqrt {a_c^2 + a_t^2} \)

Centripetal Acceleration (a_c):

\(\therefore {a_c} = \frac{{{v^2}}}{r}\)

\( \Rightarrow {a_c} = \frac{{{{\left( {10} \right)}^2}}}{{25}} = \frac{{100}}{{25}} = 4\;m/{s^2}\)

Hence, net acceleration 

Concept:

Work done = Area under force deflection diagram

\(Work ~done = \frac 12 \times Force\times deflection\)

Calculation:

Given:

Gradual loading

Force = 100 N, deflection = 100 mm = 0.1 m

\(Work ~done = \frac 12 \times Force\times deflection\)

\(Work ~done = \frac 12 \times 100\times 0.1=5~J\)

Explanation:

When the Force F is suddenly removed, then due to W, the rod is in rotating condition with angular acceleration \(\alpha\)

Thus the equation of motion:

\(\Sigma M=I_o\alpha;\;\;W \times \frac{L}{2} = I\alpha\)

\(As, I = \frac{mL^2}{3}= \frac{1}{3}\times\frac{W}{g}\times{L^2}\)

\(\therefore W \times \frac{L}{2} =\frac{WL^2}{3g}\times\alpha\)

\(\Rightarrow \alpha = \frac{{3g}}{{2L}}\)

\(\therefore Linear\;acceleration\;at\;centre = \alpha \times \frac{L}{2} = \frac{{3g}}{{2L}}\times \frac{L}{2}=\frac{{3g}}{4}\)

Also, the centre of the rod accelerates with linear acceleration a;

\(W-R=F\Rightarrow mg-R=ma\\R=mg-ma=mg-\frac{3}{4}mg=\frac{1}{4}mg=\frac{W}{4}\)   

Explanation:

Non-conservative forces:

• A non-conservative force is one for which work depends on the path taken.
• Friction is an example of a non-conservative force that changes mechanical energy into thermal energy.
• Work Wnc done by a non-conservative force changes the mechanical energy of a system.
• In equation form, Wnc = ΔKE + ΔPE or, equivalently, KEi + PEi = Wnc + KEf + PEf.

Conservative forces:

• A conservative force is one, for which work done by or against it depends only on the starting and ending points of a motion and not on the path taken. 
• Example potential energy, gravitational force etc.
• Work done by conservative force is equal to decrease in potential energy.

Concept:

A body moving in a horizontal circle has only kinetic energy stored in it.

Centripetal force = \(\frac{mv^2}{r}\)

where m is mass, v is the velocity of particle, and r is the radius.

The kinetic energy of a body is given by:

\(K.E=\frac{1}{2}mv^2\)

Calculation:

Given:

Centripetal force = \(-\frac{C}{r^2}\)

Centripetal force = \(\frac{mv^2}{r}\)

\(\frac{mv^2}{r}=-\frac{C}{r^2}\)

\({mv^2}{}=-\frac{C}{r}\)

\(K.E=\frac{1}{2}mv^2=\frac{1}{2}\times \left(-\frac{C}{r}\right)=-\frac{C}{2r}\)

Don't Confuse with the negative sign in the value of K.E as constant C must be negative here so that K.E become positive. 

Concept:

Absolute velocity is given by:

\(V = \sqrt {v_x^2 + v_y^2}\)

v_x = velocity component in x direction = dx/dt

v_y = velocity component in y direction = dy/dt

Calculation:

At t = 6 sec

x = 4 – 9t

v_x = dx/dt = -9

y = t²

v_y = dy/dt = 2t = 2(6) = 12

\(V = \sqrt {{{\left( { - 9} \right)}^2} + {{\left( {12} \right)}^2}} = \sqrt {225} = 15\;m/s\)

Explanation:

Acceleration:

• The object under motion can undergo a change in its speed. The measure of the rate of change in its speed along with direction with respect to time is called acceleration.
• The motion of the object can be linear or circular.

Linear acceleration:

• The acceleration involved in linear motion is called linear acceleration.
• Here the acceleration is only due to the change in speed and no acceleration due to change in direction, therefore no radial component of acceleration.

Circular acceleration:

• The acceleration involved in a circular motion is called angular acceleration.
• In a circular motion, the acceleration experienced by the body towards the centre is called the centripetal acceleration which can be resolved into two-component.
• A radial component and a tangential component depending upon the type of motion.

Radial acceleration (a_r):

• The acceleration of the object along the radius, directed towards the centre is called radial acceleration.

\(a_r=\frac{v^2}{r}=\omega^2r\)

Tangential acceleration (a_t):

• The tangential component is defined as the component of angular acceleration tangential to the circular path.

\(a_t=\alpha r\)

Concept:

Projectile motion: 

When a particle is projected obliquely near the earth's surface, it moves simultaneously in horizontal and vertical directions. The path of such a particle is called projectile and the motion is called projectile motion.

Range of projectile:

• The horizontal range of a projectile is the distance along the horizontal plane it would travel, before reaching the same vertical position as it started from.

Formulae in projectile motion:

\(Range\;of\;projectile = \frac{{{u^2}\sin 2θ }}{g}\)

\(Total\;time\;of\;flight = \frac{{2u\;sinθ }}{g}\)

\(Maximum\;Height = \frac{{{u^2}{{\sin }^2}θ }}{{2g}}\)

where u = projected speed, θ = angle at which an object is thrown from the ground and g = acceleration due to gravity = 9.8 m/s².

Calculation:

Given:

Range of a Projectile motion is given by (R):

\(R= \frac{{{u^2}\sin 2θ }}{g}\)

For horizontal distance to be maximum:

sin 2θ = 1

∴ sin 2θ = sin 90° 

∴ θ = 45°.

∴ to take the longest possible jump, an athlete should make an angle of 45° with the ground.