Physics Chapter Test
Topic: Work, Energy, Power and Conservation Laws (45 Questions)
Q.1 If the values of force and length are increased four times then the unit of energy will increase by–
Correct Answer: (4) 16 times
Explanation: Energy = Force × Displacement. If both Force and Length (displacement) are increased 4 times, Energy increases by 4 × 4 = 16 times.
Q.2 A bullet of mass P is fired with velocity Q in a large body of mass R. The final velocity of the system will be:
Correct Answer: (2) PQ / (P + R)
Explanation: By conservation of momentum, P × Q = (P + R) × V, hence the final velocity of the system V = PQ / (P + R).
Q.3 A sphere of mass m moving with a constant velocity collides with another stationary sphere of same mass. The ratio of velocities of two spheres after collision will be, if the co-efficient of restitution is e:
Correct Answer: (1) (1 - e) / (1 + e)
Explanation: Using momentum conservation: m*u = m*v1 + m*v2 => v1 + v2 = u. From definition of coefficient of restitution: e*u = v2 - v1. Solving for v1 and v2 yields v1 = u*(1-e)/2 and v2 = u*(1+e)/2. Thus, the ratio v1/v2 = (1 - e) / (1 + e).
Q.4 An electric motor produces a tension of 4500 N in a load lifting cable and rolls it at the rate of 2 m/s. The power of the motor is –
Correct Answer: (1) 9 KW
Explanation: Power = Force × Velocity = 4500 N × 2 m/s = 9000 W = 9 kW.
Q.5 A ball falls from a height of 5 m and strikes the roof of a lift. If at the time of collision, lift is moving in the upward direction with a velocity of 1 m/s, then the velocity with which the ball rebounds after collision will be – (e = 1)
Correct Answer: (2) 12 m/s upwards
Explanation: Velocity of ball before collision v = √(2gh) = √(2 × 10 × 5) = 10 m/s downwards. Relative velocity of approach = 10 + 1 = 11 m/s. Since e = 1, relative velocity of separation is also 11 m/s. Velocity of ball after collision v' - 1 = 11 => v' = 12 m/s upwards.
Q.6 A force F = 5i + 3j N is applied over a particle which displaces it from its origin to the point r = 2i - j m. The work done on the particle in joules is:
Correct Answer: (2) + 7
Explanation: Work Done = F • r = (5i + 3j) • (2i - j) = 5(2) + 3(-1) = 10 - 3 = 7 J.
Q.7 Two elastic bodies P and Q having equal masses are moving along the same line with velocities of 16 m/s and 10 m/s respectively. Their velocities after the elastic collision will be in m/s:
Correct Answer: (3) 10 and 16
Explanation: When two elastic bodies of equal masses undergo head-on elastic collision, they interchange their velocities. Therefore, the velocities after collision will be 10 m/s and 16 m/s.
Q.8 If the momentum of a body is increased n times, its kinetic energy increases:
Correct Answer: (4) n2 times
Explanation: Kinetic Energy K = p2 / 2m. If momentum p is increased n times (p' = np), then K' = (np)2 / 2m = n2 × K. Thus, kinetic energy increases by n2 times.
Q.9 A metal ball does not rebound when struck on a wall, whereas a rubber ball of same mass when thrown with the same velocity on the wall rebounds. From this it is inferred that –
Correct Answer: (2) Change in momentum in rubber ball is more
Explanation: For the metal ball, final velocity is 0, so change in momentum is m*v. For the rubber ball, it rebounds with some velocity (say, -v), so change in momentum is m*v - (-m*v) = 2m*v (which is larger).
Q.10 The unit of the co–efficient of restitution is –
Correct Answer: (4) None of the above
Explanation: The coefficient of restitution is the ratio of relative velocity of separation to relative velocity of approach. It is a ratio of identical quantities and is therefore dimensionless and unitless.
Q.11 A bomb of mass 9 kg explodes into two pieces of 3 kg and 6 kg. The velocity of 3 kg piece is 16 m/s. The kinetic energy of 6 kg piece is –
Correct Answer: (3) 192 Joule
Explanation: By conservation of momentum: m1*v1 = m2*v2 => 3 × 16 = 6 × v => v = 8 m/s. Kinetic energy of 6 kg piece = 1/2 × m2 × v22 = 1/2 × 6 × 82 = 3 × 64 = 192 J.
Q.12 Two solid rubber balls A and B whose masses are 200 gm and 400 gm respectively, are moving in mutually opposite directions. If the velocity of A is 0.3 m/s and both the balls come to rest after collision, then the velocity of ball B is –
Correct Answer: (2) – 0.15 m/s
Explanation: By momentum conservation: m_A*v_A + m_B*v_B = 0 => (0.2 kg)*(0.3 m/s) + (0.4 kg)*v_B = 0 => 0.06 + 0.4*v_B = 0 => v_B = -0.15 m/s.
Q.13 The graph between p / EK and p is (EK = kinetic energy and p = momentum) –
Correct Answer: (3) Rectangular hyperbola
Explanation: Since EK = p2 / 2m, we have p / EK = 2m / p. Let y = p / EK and x = p. Then y = 2m / x => y × x = 2m = constant. This represents a rectangular hyperbola.
Q.14 A 1 kg ball falls from a height of 25 cm and rebounds up to a height of 9 cm. The coefficient of restitution is:
Correct Answer: (1) 0.6
Explanation: Coefficient of restitution e = √(h_rebound / h_initial) = √(9 / 25) = 3 / 5 = 0.6.
Q.15 The graph between potential energy U and displacement X in the state of stable equilibrium will be–
Correct Answer: (1) A curve with a minimum (bowl-shaped opening upwards)
Explanation: In a state of stable equilibrium, the potential energy (U) is at a minimum value. Therefore, the graph of U vs X shows a curve with a minimum (bowl-shaped opening upwards).
Q.16 A force F = (3x2 + 2x – 7) N acts on a 2 kg body as a result of which the body gets displaced from x = 0 to x = 5 m. The work done by the force will be –
Correct Answer: (3) 115 Joule
Explanation: Work Done = ∫ F dx from 0 to 5 = ∫ (3x2 + 2x – 7) dx = [x3 + x2 – 7x] from 0 to 5 = (125 + 25 - 35) - 0 = 115 J.
Q.17 A 50 gm bullet moving with a velocity of 10 m/s gets embedded into a 950 gm stationary body. The loss in kinetic energy of the system will be –
Correct Answer: (4) 95%
Explanation: Initial KE K1 = 1/2*m*v2. Final velocity after embedding: V = m*v / (m+M) = 50*10 / 1000 = 0.5 m/s. Final KE K2 = 1/2*(m+M)*V2 = 1/2 * 1 * (0.5)2 = 0.125 J. Initial KE K1 = 1/2 * 0.05 * 102 = 2.5 J. Loss in KE % = (K1 - K2)/K1 × 100 = (2.5 - 0.125)/2.5 × 100 = 95%.
Q.18 A crane lifts 300 kg weight from earth's surface up to a height of 2 m in 3 seconds. The average power generated by it will be –
Correct Answer: (1) 1960 W
Explanation: Work Done = mgh = 300 × 9.8 × 2 = 5880 J. Power = Work / Time = 5880 / 3 = 1960 W.
Q.19 A body is dropped from a height h. When loss in its potential energy is U then its velocity is v. The mass of the body is –
Correct Answer: (4) 2U / v2
Explanation: By conservation of energy, Loss in PE = Gain in KE => U = 1/2 × m × v2 => m = 2U / v2.
Q.20 A block of mass 16 kg is moving on a frictionless horizontal surface with velocity 4 m/s and comes to rest after pressing a spring. If the force constant of the spring is 100 N/m then the compression in the spring will be –
Correct Answer: (2) 1.6 m
Explanation: By conservation of energy: 1/2 × m × v2 = 1/2 × k × x2 => 16 × 42 = 100 × x2 => 256 = 100 x2 => x2 = 2.56 => x = 1.6 m.
Q.21 A bomb initially at rest explodes by itself into three equal mass fragments. The velocities of two fragments are (3i + 2j) m/s and (–i – 4j) m/s. Velocity of the third fragment is (in m/s):
Correct Answer: (3) -2i + 2j
Explanation: By conservation of momentum, total momentum is conserved: m*v1 + m*v2 + m*v3 = 0 => v3 = -(v1 + v2) = -((3i + 2j) + (-i - 4j)) = -(2i - 2j) = -2i + 2j.
Q.22 A ball of mass 1 kg is released from the tower of Pisa. The kinetic energy generated in it after falling through 10 m will be –
Correct Answer: (4) 98 J
Explanation: Kinetic Energy generated = Potential Energy lost = mgh = 1 × 9.8 × 10 = 98 J.
Q.23 A 10 kg satellite completes one revolution around the earth at a height of 100 km in 108 minutes. The work done by the gravitational force of earth will be –
Correct Answer: (3) 0 J
Explanation: The gravitational force on the satellite acts as a centripetal force which is always perpendicular to the direction of motion (displacement). Hence, the work done is zero (W = F • ds = F ds cos(90°) = 0).
Q.24 A particle moves in a potential region given by U = 8x2 – 4x + 400 J. Its state of equilibrium will be –
Correct Answer: (2) x = 0.25 m
Explanation: At equilibrium, F = -dU/dx = 0 => d(8x2 - 4x + 400)/dx = 0 => 16x - 4 = 0 => x = 0.25 m.
Q.25 A person of mass m is standing on one end of a plank of mass M and length L and floating in water. The person moves from one end to another and stops. The displacement of the plank is –
Correct Answer: (1) mL / (M + m)
Explanation: Since there is no external horizontal force on the system, the center of mass remains stationary. If the person moves by L relative to the plank and the plank shifts by x relative to the water, m(L - x) = Mx => mL = (M + m)x => x = mL / (M + m).
Q.26 A bullet of mass m moving with a speed v strikes a wooden block of mass M and gets embedded into the block. The final speed is:
Correct Answer: (3) m v / (M + m)
Explanation: By conservation of linear momentum: m*v = (M + m)*U => final speed U = mv / (M + m).
Q.27 Two men with weights in the ratio 5 : 3 run up a staircase in times in the ratio 11 : 9. The ratio of power of first to that of second is –
Correct Answer: (1) 15 / 11
Explanation: Power P = Work / Time = mgh / t. Ratio P1 / P2 = (w1/w2) × (t2/t1) = (5/3) × (9/11) = 15/11.
Q.28 The retarding force required to reduce velocity of a 3 kg body from 0.75 m/s to 0.25 m/s in 0.02 sec will be –
Correct Answer: (3) 75 N
Explanation: Acceleration a = (v - u) / t = (0.25 - 0.75) / 0.02 = -0.50 / 0.02 = -25 m/s2. Retarding force = m × |a| = 3 kg × 25 m/s2 = 75 N.
Q.29 A 2 kg mass lying on a table is displaced in the horizontal direction through 50 cm. The work done by the normal reaction will be –
Correct Answer: (1) 0
Explanation: The normal reaction force is directed vertically upwards, while the displacement is purely horizontal. Since the angle between them is 90°, Work Done = F ds cos(90°) = 0.
Q.30 A car is moving with a speed of 40 km/hr. If the car engine generates 7 kilowatt power, then the resistance in the path of motion of the car will be –
Correct Answer: (2) 630 Newton
Explanation: Velocity v = 40 km/hr = 40 × (5/18) = 100/9 m/s. Power P = F × v => 7000 W = F × (100/9) => F = 70 × 9 = 630 N.
Q.31 Figure shows the vertical section of a frictionless surface. A block of mass 2 kg is released from the position A (height = 10 m); its kinetic energy as it reaches the position C (height = 3 m) is:
Correct Answer: (2) 140 J
Explanation: By conservation of mechanical energy: Loss in PE = Gain in KE => KE = mg(h_A - h_C) = 2 × 10 × (10 - 3) = 2 × 10 × 7 = 140 J.
Q.32 A force F = Kx2 acts on a particle at an angle of 60° with the x–axis. The work done in displacing the particle from x1 to x2 will be –
Correct Answer: (3) 1/6 K (x23 - x13)
Explanation: dW = F dx cos(60°) = Kx2 dx × 0.5. Integrating from x1 to x2: W = 0.5 × K [x3/3] from x1 to x2 = 1/6 K (x23 - x13).
Q.33 A ball moving with velocity of 9 m/s collides with another similar stationary ball. After the collision both the balls move in directions making an angle of 30° with the initial direction. After the collision their speed will be:
Correct Answer: (2) 5.2 m/s
Explanation: By conservation of momentum along the initial direction: m*u = m*v1*cos(30°) + m*v2*cos(30°). Since both balls are similar and move at the same angle, their velocities after collision are equal: v1 = v2 = v. Thus, u = 2v*cos(30°) => 9 = 2v*(√3/2) = v*√3 => v = 9 / √3 = 3√3 ≈ 5.2 m/s.
Q.34 A bomb of 50 kg is fired from a cannon with a velocity 600 m/s. If the mass of the cannon is 103 kg, then its recoil velocity will be –
Correct Answer: (2) –30 m/s
Explanation: By conservation of momentum: m*v_shell + M*v_recoil = 0 => 50 × 600 + 1000 × v_recoil = 0 => v_recoil = -30 m/s.
Q.35 Two masses m1 = 2 kg and m2 = 5 kg are moving on a frictionless surface with velocities 10 m/s and 3 m/s respectively. m2 is ahead of m1. An ideal spring of spring constant k = 1120 N/m is attached on the back side of m2. The maximum compression of the spring will be:
Correct Answer: (3) 0.25 m
Explanation: Maximum compression happens when both have the same velocity. Common velocity V = (m1*u1 + m2*u2)/(m1 + m2) = (2*10 + 5*3)/(2+5) = 35/7 = 5 m/s. Energy conservation: 1/2*m1*u12 + 1/2*m2*u22 = 1/2*(m1+m2)*V2 + 1/2*k*x2 => 1/2*2*100 + 1/2*5*9 = 1/2*7*25 + 1/2*1120*x2 => 100 + 22.5 = 87.5 + 560*x2 => 35 = 560*x2 => x2 = 35/560 = 1/16 => x = 1/4 = 0.25 m.
Q.36 A solid sphere is moving and it makes an elastic collision with another stationary sphere of half of its own radius. After collision it comes to rest. The ratio of the densities of materials of second sphere and first sphere is –
Correct Answer: (3) 8
Explanation: Since the first sphere comes to rest after elastic collision with a stationary sphere, their masses must be equal: m1 = m2 => ρ1 × 4/3 π r13 = ρ2 × 4/3 π r23 => ρ2 / ρ1 = (r1 / r2)3. Given r2 = r1 / 2, we have ρ2 / ρ1 = (2)3 = 8.
Q.37 The mass of a bucket full of water is 15 kg. It is being pulled up from a 15 m deep well. Due to a hole in the bucket 6 kg water flows out of the bucket. The work done in drawing the bucket out of the well will be –
Correct Answer: (3) 1800 Joule
Explanation: Initial mass is 15 kg, final mass is 15 - 6 = 9 kg. Since water leaks at a uniform rate, the average mass pulled is (15 + 9)/2 = 12 kg. Work Done = m_avg × g × h = 12 kg × 10 m/s2 × 15 m = 1800 J.
Q.38 A 5 kg body collides with another stationary body. After the collision, the bodies move in the same direction with one–third of the velocity of the first body. The mass of the second body will be –
Correct Answer: (2) 10 kg
Explanation: Initial momentum = m1 × u1 = 5 × u1. After collision, both move with velocity u1/3 in the same direction. By conservation of momentum: 5 × u1 = (5 + m2) × (u1/3) => 15 = 5 + m2 => m2 = 10 kg.
Q.39 A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac = k2 r t2, where k is a constant. The power delivered to the particle by the forces acting on it will be –
Correct Answer: (4) m k2 r2 t
Explanation: ac = v2/r = k2 r t2 => v = k r t. Tangential acceleration at = dv/dt = k r. Tangential force Ft = m*at = m*k*r. Power delivered P = Ft × v = (m*k*r) × (k*r*t) = m k2 r2 t.
Q.40 A 10 g bullet, moving with a velocity of 500 m/s, enters a stationary piece of ice of mass 10 kg and stops. If the piece of ice is lying on a frictionless plane, then its velocity will be:
Correct Answer: (3) 0.5 m/s
Explanation: By conservation of momentum: m*u = (M + m)*v => (0.01 kg) × 500 m/s = (10 + 0.01) kg × v => 5 ≈ 10*v => v ≈ 0.5 m/s.
Q.41 A 5 gm lump of clay, moving with a velocity of 10 cm/s towards east, collides head–on with another 2 gm lump of clay moving with 15 cm/s towards west. After collision, the two lumps stick together. The velocity of the compound lump will be –
Correct Answer: (3) 2.88 cm/s towards east
Explanation: Let East be positive direction. Initial momentum = 5 × 10 + 2 × (-15) = 50 - 30 = +20 gm cm/s. Combined mass = 5 + 2 = 7 gm. Final velocity v = 20 / 7 ≈ 2.86 cm/s (option 2.88 cm/s) towards east.
Q.42 A frictionless steel ball of radius 2 cm, moving on a horizontal plane with a velocity of 5 cm/s, collides head–on with another stationary steel ball of radius 3 cm. The velocities of two bodies after collision will respectively be (in cm/s) (e = 1) –
Correct Answer: (2) –2.7, 2.3
Explanation: Mass is proportional to volume, so m ∝ r3. m1 ∝ 23 = 8 and m2 ∝ 33 = 27. Let m1 = 8k, m2 = 27k. For head-on elastic collision with stationary target: v1 = (m1 - m2)/(m1 + m2)*u1 = (8 - 27)/35 × 5 = -19/7 ≈ -2.7 cm/s. v2 = 2*m1/(m1 + m2)*u1 = 16/35 × 5 = 16/7 ≈ 2.3 cm/s.
Q.43 A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by x = 3t – 4t2 + t3, where x is in metres and t is in seconds. The work done during the first 4 second is:
Correct Answer: (1) 5.28 J
Explanation: v = dx/dt = 3 - 8t + 3t2. At t = 0, v1 = 3 m/s. At t = 4, v2 = 3 - 32 + 48 = 19 m/s. Work Done = change in KE = 1/2 × m × (v22 - v12) = 1/2 × 0.03 × (192 - 32) = 0.015 × (361 - 9) = 0.015 × 352 = 5.28 J.
Q.44 A man is supplying an instantaneous power of 500 J/s to a massless string by pulling it at an instantaneous speed of 10 m/s. It is known that kinetic energy of the block is increasing at a rate of 100 J/s at that instant. Then the mass of the block is –
Correct Answer: (4) 4 kg
Explanation: Power supplied to string P = F × v => 500 = F × 10 => Tension T = F = 50 N. The rate of increase of KE of block is Net Force × velocity => (T - mg) × v = 100 => (50 - m × 10) × 10 = 100 => 50 - 10m = 10 => 10m = 40 => m = 4 kg.
Q.45 A shell is fired from a cannon with velocity V m/s at an angle θ with the horizontal direction. At the highest point in its path with same speed it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon. The speed in m/sec. of the other piece immediately after the explosion is:
Correct Answer: (2) 3V cos θ
Explanation: At the highest point, velocity is V cos θ horizontal. Initial momentum = 2m × V cos θ. After explosion: one piece retraces its path with velocity -V cos θ. Let the velocity of the other piece be V'. By momentum conservation: 2m × V cos θ = m(-V cos θ) + m*V' => 2V cos θ = -V cos θ + V' => V' = 3V cos θ.