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QUESTION BANK TEST#1 CHAP#2 FORCE & MOTION
2.3 PRODUCT OF TWO VECTORS
DOT & CROSS PRODUCT
1. The dot product of two perpendicular vectors is: A) Maximum B) Zero C) Minimum D) Negative 2. The dot product of two vectors A and B is given by: A) ( AB sin θ ) B) ( AB cos θ) C) ( A + B ) D) ( A - B ) 3. If (θ= 0^0 ) between two vectors, their dot product is: A) Zero B) Minimum C) Maximum D) Negative 4. The dot product of a vector with itself is equal to: A) Zero B) Vector quantity C) Square of its magnitude D) Its direction 5. Which of the following is a scalar quantity obtained using dot product? A) Torque B) Angular momentum C) Work D) Force 6. If (A.B = 0) then the vectors are: A) Parallel B) Anti-parallel C) Perpendicular D) Equal 7. The unit of dot product of force and displacement is: A) Newton B) Joule C) Watt D) Pascal 8. The dot product of two vectors depends on: A) Only magnitudes B) Only angle between them C) Magnitudes and angle between them D) Directions only Here are 8 MCQs on Cross (Vector) Product suitable for intermediate / Punjab Board Physics , with answers at the end: 9. The cross product of two vectors is a: A) Scalar quantity B) Vector quantity C) Dimensionless quantity D) Tensor quantity 10. The magnitude of cross product of two vectors A and B is: A) AB cos θ B) AB tan θ C) AB sin θ D) ( A + B ) 11. The direction of cross product is given by: A) Left hand rule B) Fleming’s right hand rule C) Right hand thumb rule D) Corkscrew rule
12. The cross product of two parallel vectors is: A) Maximum B) Zero C) Minimum D) Negative 13. Which physical quantity is obtained using cross product? A) Work B) Power C) Energy D) Torque 14. If the angle between two vectors is (90^0 ), the magnitude of their cross product is: A) Zero B) Minimum C) Maximum D) Negative 15. The unit of cross product of force and displacement is: A) Joule B) Newton C) Newton–meter D) Watt 16. The cross product of a vector with itself is: A) Maximum B) Minimum C) Zero D) One **SHORT QUESTIONS
- Prove that vector is a unit vector or find its magnitude? Answer:** As we know that we can find the magnitude of a vector as A = A = A = 1 (A is a unit vector because unit vector has magnitude “1”) 2. If all the components of the vectors, were reversed, how would this alter? Answer: If all the components of the vectors are reversed, then we get new vectors which are negative of i.e Now It shows that will not alter. 3. Name the three different conditions that could make Ans: (i) Either of vectors is null vector.
Rotate the first vector into through the smaller of two possible angles. This rotation is represented by curling the fingers of stretched right hand placed on the first vector , then thumb represents the direction of vector product. The direction of x will be vertically upward as shown in the fig (A). According to this direction rule, x is a vector opposite to the direction of as shown in fig (B). Hence x = - x
8. Prove that Ans: Let we have then **2.4: EQUATIONS OF MOTIONS (without graphical methods)
- Which of the following is the first equation of motion?** A) v = vi + at B) s = vit + 1/2 at^2 ) C) v^2 = vi^2 + 2as D) a = vi + vf 2. A body starts from rest and accelerates uniformly at 2 m/s². What will be its velocity after 5 s? A) 5 m/s B) 8 m/s C) 10 m/s D) 12 m/s 3. If initial velocity is zero, which equation is most suitable to find distance traveled? A) v = vi + at B) s = vit + 1/2 at^2 ) C) v^2 = vi^2 + 2as D) a = vi + vf 4. A car moving with uniform velocity has acceleration: A) Zero B) Increasing C) Decreasing D) Infinite 5. A body comes to rest after 10 s with a uniform deceleration of 2 m/s². What was its initial velocity? A) 10 m/s B) 15 m/s C) 20 m/s D) 25 m/s 6. Which equation of motion does NOT contain time? A) v = vi + at B) s = vit + 1/2 at^2 ) C) v^2 = vi^2 + 2as D) All of these
7. A body moving with acceleration 4 m/s² has initial velocity 6 m/s. What is its velocity after 3 s? A) 12 m/s B) 15 m/s C) 18 m/s D) 20 m/s 8. A body is thrown vertically upward. Which quantity remains constant? A) Velocity B) Displacement C) Acceleration D) Momentum 9. Which condition is necessary for using equations of motion? A) Variable acceleration B) Uniform acceleration C) Zero velocity D) Circular motion SHORT QUESTIONS Q1. Write the first equation of motion and explain its terms. Ans: The first equation of motion is vf = vi + at. Here “ vi ” is initial velocity, “ vf” is final velocity, “ a ” is acceleration, and “ t ” is time. Q2. State the second equation of motion. Ans: The second equation of motion is s = vi t + ½ at². It gives the distance s covered by a body moving with uniform acceleration. Q3. Write the third equation of motion. Ans: The third equation of motion is vi² - vi^2 = 2as. It relates velocity with acceleration and distance without involving time. Q4. A body starts from rest and accelerates at 2 m/s² for 5 s. Find its final velocity. Ans: Using v = vi + at , where vi = 0. v = 0 + (2 × 5) = 10 m/s. Q5. Define uniform acceleration. Ans: Uniform acceleration is when the velocity of a body changes by equal amounts in equal intervals of time. Q6. When can equations of motion be applied? Ans: Equations of motion are applied when acceleration is constant and motion is in a straight line. 2.5 MOTION UNDER GRAVITY
MCQs: Motion Under Gravity
6. What is the velocity of a body at the highest point of its upward motion? Answer: At the highest point, the velocity of the body becomes zero. However, the acceleration due to gravity is still 9.8 m/s² downward. **2.6 PROJECTILE MOTION
- A projectile is launched at an angle of 30° with the horizontal with a speed of 20 m/s. The horizontal range is:** A) 20 m B) 30 m C) 35.3 m D) 40 m 2. At the highest point of its trajectory, the velocity of a projectile is: A) Zero B) Entirely horizontal C) Entirely vertical D) Both horizontal and vertical components 3. The time of flight of a projectile depends on: A) Initial speed only B) Angle of projection only C) Both initial speed and angle of projection D) Mass of the projectile 4. If the angle of projection of a projectile is 45°, the range will be: A) Maximum B) Minimum C) Half of the maximum D) Independent of initial speed 5. The vertical and horizontal motions of a projectile are: A) Dependent on each other B) Completely independent of each other C) Only horizontal affects vertical D) Only vertical affects horizontal 6. A projectile is fired horizontally from the top of a cliff. Which statement is true about its motion? A) Horizontal velocity decreases B) Vertical acceleration is zero C) Horizontal velocity remains constant D) Vertical velocity is constant 7. A projectile is launched with speed (vi) at an angle (θ). Its maximum height (H) is given by: A) vi^2 sin^2 θ/2gB) vi^2 cos^2 θ/2g C) vi^2 /2g D) vi^2 sin2θ/2g 8. For a projectile, the path is: A) A straight line B) A parabola C) A circle D) An ellipse 9. Two projectiles are launched with the same speed but at complementary angles (30° and 60°). Which of the following is true? A) They have the same range B) They have the same maximum height C) They have the same time of flight D) Both B and C 10. The horizontal component of velocity of a projectile: A) Decreases uniformly B) Increases uniformly C) Remains constant D) Becomes zero at maximum height
SHORT QUESTIONS
1. What is projectile motion? Answer: Projectile motion is the motion of an object thrown into the air under the influence of gravity alone. The object follows a curved (parabolic) path. Air resistance is neglected in ideal projectile motion. 2. Why is the path of a projectile parabolic? Answer: Horizontal motion occurs with constant velocity, while vertical motion has constant acceleration due to gravity. These two independent motions combine to form a parabola. This shape is a result of uniform acceleration in one direction. 3. What is the time of flight of a projectile? Answer: Time of flight is the total time the projectile remains in the air. It depends on the vertical component of velocity and gravity. Greater initial vertical velocity increases the time of flight. T = 2vi sin θ/g = Tascending + Tdescending 4. What factors affect the range of a projectile? Answer: The range depends on the initial velocity, angle of projection, and gravity. For a given speed, maximum range occurs at 45°. It is independent of the mass of the projectile. R = vi^2 **sin 2θ/g
- Why is horizontal velocity constant in projectile motion? Answer:** No force acts on the projectile in the horizontal direction (air resistance ignored). Therefore, there is no horizontal acceleration. As a result, horizontal velocity remains constant throughout the motion. 2.7 MOMENTUM (1) The product of force and time is called (A) Impulse (B) momentum (C) acceleration (D) torque (2) A force of 10 N act on a body of mass 10 kg for 5 sec. The change in momentum of the body is (A) 10 Ns (B) 100 Ns (C) 50 Ns (D) 150 Ns (3) Force acting on a body is equal to (A) rate of change of momentum (B) change of momentum (C) momentum (D) both A&B (4) Dimension of momentum is (A) [MLT] (B) [MLT^2 ] (C) [ML-1T] (D) [MLT-1] (5) The unit of impulse is (A) Nm (B) Ns (C) Ns-1^ (D) N
5. Why does a heavier object have more momentum than a lighter one at the same speed? Answer: Momentum depends directly on mass. For the same velocity, a heavier object has greater mass. Hence, it possesses greater linear momentum. ( p = mv ) ↔ p ∝ m v = constant 6. Give one daily life example of conservation of momentum. Answer: 1. When a gun is fired , the bullet moves forward and the gun recoils backward. The forward momentum of the bullet equals the backward momentum of the gun. Total momentum of the system remains conserved. 2. When a person jumps off a boat , the person moves forward while the boat moves backward. The forward momentum of the person is balanced by the backward momentum of the boat. Thus, the total momentum of the system remains conserved. 7. What is impulse? Answer: Impulse is the product of force and the time for which it acts. It is equal to the change in linear momentum of an object. Mathematically, impulse ( F ∆t = ∆p ) 8. Why does a cricketer pull his hands back while catching a ball? Answer: Pulling the hands back increases the time of contact. This reduces the force experienced by the hands. Thus, the ball is caught safely without injury. 9. A force of 10 N acts on a body for 0.2 s. Find the impulse produced or change in momentum? Answer: Impulse is given by the relation ( F ∆t = ∆p ). Impulse = 10 × 0. The impulse produced is 2 N·s. **2.8 Elastic Collision and In-elastic Collision (in one-dimension)
- The system on which no external force acts is called** (A) isolated system (B) open system (C) non-inertial system (D) thermal system 2. The collision in which linear momentum as well as K.E is conserved is called (A) nearlyelastic collision (B) perfectly elastic collision (C) nonelastic collision (D) none of these 3. In an inelastic collision (A) momentum is conserved (B) total energy is conserved (C) both a & b (D) both are not conserved
4. In case of elastic collision (A) magnitude of relative speed of approach equal to the magnitude of relative speed of separation (B) magnitude of relative speed of approach is doubled of the magnitude of relative speed of separation (C) magnitude of relative speed of approach greater to magnitude of relative speed of separation (D) magnitude of relative speed of approach very less to the magnitude of relative speed of separation 5. For two colliding balls which condition is applicable for one dimensional elastic collision (A) they should be non-rotating (B) they should be smooth (C) both A and B (D) they should not be smooth 6. When a very heavy ball ‘B 1 ’ collide with a stationary target ‘B 2 ’ of negligible mass, after collision the final velocity of ball ‘B 2 ’ will (A) become zero (B) become half (C) become doubled as compared to B 1 (D) same as the B 1 7. When two objects undergo an inelastic collision then (A) objects come to rest after collision (B) momentum of the objects changes (C) momentum does not change (D) the law of conservation of energy is violated 8. In a perfectly elastic collision, which quantity is conserved? A. Only momentum B. Only kinetic energy C. Momentum and kinetic energy D. Momentum and potential energy 9. Two bodies collide elastically. After collision, the total kinetic energy of the system is: A. Greater than before collision B. Less than before collision C. Equal to before collision D. Zero 10. Which of the following is an example of an approximately elastic collision? A. Car crash B. Clay sticking to a wall C. Collision between billiard balls D. Bullet embedding in a block 11. In a one-dimensional elastic collision between two equal masses, what happens to their velocities after collision? A. Both come to rest B. They stick together C. One moves faster than before D. They exchange velocities SHORT QUESTIONS
2.9 Elastic collision involves: a) loss of energy b) gain of energy c) no gain, no loss of energy d) no relation between energy and elastic collision SHORT ANSWER QUESTIONS Q#2.1 : State right hand rule for two vectors with respect to vector product. Answer Right Hand Rule The right-hand rule is used to determine the direction of the vector product (cross product) of two vectors. Explanation If you point your right hand’s fingers in the direction of the first vector ( A ) and curl them toward the second vector ( B ), your thumb points in the direction of A × B. Direction of A × B is perpendicular to the plane containing A and B , following the right-hand rule. Q# 2.2: Define impulse and show how it is related to momentum. Answer: Definition “ When large force acts on a body for a short interval of time and changes its momentum, the product of such force and time is called Impulse ”. It is a vector quantity Mathematically; Impulse = Force time = t = t = mvf - mvi Relation with Momentum: Whenever the impulse is generated in a body its linear momentum changes. Means Impulse equals the change in momentum of the body. Impulse = Change in momentum = mvf – mvi Q# 2.3: Differentiate between an elastic and an inelastic collision. Answer: Elastic Collision: In case of elastic collision, both momentum and kinetic energy are conserved. ∆K.Ei = ∆K.Ef ∆Pi = ∆Pf Inelastic Collision: In case of an inelastic collision Momentum is conserved, but kinetic energy is not. K.E loss because of friction. ∆K.Ei ≠ ∆K.Ef ∆Pi = ∆Pf
Q# 2.4: Show that rate of change in momentum is equal to force applied. Also state Newton's second law of motion in terms of momentum. Answer: Newton’s Second Law of Motion in Terms of Momentum: Newton’s second law of motion can also be stated in terms of momentum as follows: “ Time rate of change of momentum of a body equals the applied force ” The acceleration produced by this force is given by; = (i) By Newton’s second law, = m = (ii) Equating the two expressions of acceleration, we have; = ↔ F = ∆P/∆t Q# 2.5: State law of conservation of linear momentum. Also state condition under which it holds. Answer: Statement “ Total linear momentum of an isolated system remains constant (if no external force acts on it) ” Condition System must be isolated; net external force = 0 or in actual case, changes in system must be very slow or external forces are very small. m u₁ ₁ + m u₂ ₂ = m v₁ ₁ + m v₂ ₂ (in absence of external force) Q# 2.6: Show that range of projectile is maximum at an angle of 45°. Answer: As we know that the relation for range of projectile R = (vi² sin2θ) /g For maximum range Sin2θ = 1 ↔ 2θ = Sin-1^ (1) sin2θ is maximum when 2θ = 90° ⇒ θ = 45° Hence range of projectile is maximum at θ = 45°. Q# 2.8: The maximum horizontal range of a projectile is 800 m. Find the height attained by the projectile at θ = 60°. Given Data Rmax = 800 m, θ = 60°, g = 9.8 m/s² H =? Solution h = vi^2 Sin^2 θ/2g = vi^2 /g ×Sin^2 θ/2 : Rmax = vi^2 /g
Answer: Air resistance opposes the motion of a projectile as it travels through the air. Due to this resistance: The horizontal component of velocity decreases , reducing the distance the projectile can travel. The range becomes shorter compared to the ideal (vacuum) case. The trajectory is no longer symmetrical , and the projectile falls to the ground sooner. In summary, air resistance decreases the range of a projectile by slowing it down and reducing its flight time.
