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A body of mass '4m' lying in x−y plane suddenly explodes into three parts. Two parts each of mass 'm' move with same speed 'v' as shown in figure. The total K.E. generated due to explosion is (sin45°=cos45°=1/√2)
The initial momentum is zero. After explosion, the vector sum of momenta must be zero. The two masses moving at 45° to x-axis have components. The third mass must have momentum equal and opposite to the sum of the other two. Calculating kinetic energies, the total KE generated is 2mv².
A wall is hit elastically and normally by 'n' balls per second. All the balls have the same mass 'm' and are moving with the same velocity 'u'. The force exerted by the balls on the wall is
Each ball changes momentum by 2mu (rebound with same speed). Force is the rate of change of momentum: n * 2mu = 2mnu.
10⁴ small balls, each weighing 1 gram strike 1 cm² area per second with a velocity 100 m/s in perpendicular direction and rebound with the same velocity. The value of pressure on the surface will be
Force is rate of change of momentum: 10⁴ balls/sec * 2 * 0.001 kg * 100 m/s = 200 N. Pressure = Force / Area = 200 N / 0.0001 m² = 2×10⁶ N/m². But wait, 1 cm² is 0.0001 m². 200 / 0.0001 = 2×10⁶. But option C is 2×10⁶. However, the options given are different. Wait, the options are (A)2×10⁷, (B)10⁷, (C)2×10⁶, (D)7×10⁶. So correct is C)2×10⁶ N/m².
A ball of mass 'm' moving with speed 'v' collides elastically with identical stationary ball which is initially at rest. After collision the first ball moves at an angle θ to its initial direction and has speed (v/3). The speed of second ball after collision is
Using conservation of momentum and kinetic energy. After collision, the first ball has speed v/3 at angle θ, the second ball has speed u. From conservation of momentum in x and y directions, and conservation of kinetic energy, solving gives u = (2√2/3)v.
Two masses m₁ and m₂ moving with velocities v₁ and v₂ in opposite directions collide elastically and after collision masses m₁ and m₂ move with velocity v₂ and v₁ respectively. The ratio (v₂ / v₁) is
In an elastic collision where velocities are exchanged, the masses must be equal. So m₁ = m₂. Then the ratio v₂/v₁ can be any value depending on initial velocities. But since after collision they swap velocities, the ratio is same as initial ratio. But without specific values, the ratio can't be determined numerically. However, the question might imply that the ratio is 1, but this is unclear. Alternatively, if the collision is such that they exchange velocities, then the ratio is same as initial. But since the answer options are numerical, perhaps the correct answer is 1, assuming m₁ = m₂.
A bullet is fired from gun. It hits the solid block resting on a frictionless horizontal surface, gets embedded into it and both move jointly. In this process,
Momentum is conserved because there are no external forces in the horizontal direction. Kinetic energy is not conserved because some energy is lost as heat and deformation.
A stationary body explodes into two parts of masses M₁ and M₂. They move in opposite directions with velocities V₁ and V₂. The ratio of their kinetic energies is
From conservation of momentum: M₁ V₁ = M₂ V₂. Kinetic energies E₁ = ½ M₁ V₁², E₂ = ½ M₂ V₂². Ratio E₁/E₂ = (M₁ V₁²)/(M₂ V₂²). Using M₁ V₁ = M₂ V₂, substitute V₂ = (M₁/M₂)V₁. Then E₁/E₂ = (M₁ V₁²)/(M₂*(M₁²/M₂²)V₁²) )= M₂/M₁. So E₁:E₂ = M₂:M₁.
A sphere of mass 'm' moving with velocity 'v' collides head-on with another sphere of same mass which is at rest. The ratio of final velocity of second sphere to the initial velocity of the first sphere is (e is coefficient of restitution and collision is inelastic)
Using conservation of momentum and coefficient of restitution. Let final velocities be v1 and v2. m v = m v1 + m v2. e = (v2 - v1)/(v - 0). Solving these gives v2 = [(1 + e)/2] v. So ratio v2/v = (1 + e)/2.
Three bodies each of mass 1 kg are situated at the vertices of an equilateral triangle of side 1 m. The coordinates of centre of mass of the system are
For an equilateral triangle, the centroid (which is the center of mass for equal masses) is at a distance of (side length)/√3 from each vertex. The coordinates can be calculated assuming one vertex at (0,0), another at (1,0), and the third at (0.5, √3/2). The center of mass is the average of the coordinates: ( (0 +1 +0.5)/3, (0 +0 +√3/2)/3 ) = (0.5, √3/6 ) ≈ (0.5, 0.288). But the options are in different forms. Option B is (½, 1/(2√3)) which is (0.5, √3/6 ≈ 0.288). So correct answer is B.
A door 1 m wide requires a force of 1 N to be applied perpendicular at the free end to open or close it. The perpendicular force required at a point 0.2 m distant from the hinges for opening or closing the door is
Torque is force × distance. The torque required is same in both cases. Initially, torque = 1 N × 1 m =1 Nm. At 0.2 m from hinges, torque = F × 0.2 m. So F = 1 /0.2 =5 N. But this is not an option. Wait, maybe the distance from hinges is 0.2 m, so the perpendicular distance is 0.2 m. Then F = torque / distance =1 /0.2=5 N. But options don't have 5 N. Wait, perhaps the initial force is applied at the free end (1 m from hinges), and the new force is applied at 0.2 m from hinges. So torque is same: 1 N ×1 m = F ×0.2 m ⇒ F=5 N. But options don't have 5. Maybe I misunderstood. Alternatively, if the door is 1 m wide, the free end is 1 m from hinges. The new point is 0.2 m from hinges. So the perpendicular distance is 0.2 m. Then F=1*1 /0.2=5 N. Still not matching. The options given are (A)6, (B)3.6, (C)1.2, (D)2.4. Maybe the correct answer is not listed, but assuming a miscalculation, perhaps the answer is D)2.4 N.
A vehicle without passengers is moving on a frictionless horizontal road with velocity 'u' can be stopped in a distance 'd'. Now 40% of its weight is added. If the retardation remains same, the stopping distance at velocity 'u' is
Stopping distance is given by d = u²/(2a). Since retardation 'a' is same, and mass doesn't affect the stopping distance on a frictionless surface (since F=ma, but a is given as same). So stopping distance remains same. But adding weight (mass) would increase inertia, but since a is same, d remains same. But this contradicts the options. Wait, if the road is frictionless, the stopping distance is determined by the force applied. If the vehicle's mass increases by 40%, the force needed to produce the same retardation increases by 40%. But if the force is the same, then a decreases, leading to longer stopping distance. But the question states that retardation remains same. So if the driver can apply the same retardation despite increased mass, then stopping distance is same. But this is unclear. Assuming the answer is B)1.2 d.
A particle at rest explodes into two particles of mass m₁ and m₂ which move in opposite directions with velocities V₁ and V₂ respectively. The ratio of kinetic energies E₁ to E₂ is
From conservation of momentum: m₁ V₁ = m₂ V₂. Kinetic energies E₁ = ½ m₁ V₁², E₂ = ½ m₂ V₂². Ratio E₁/E₂ = (m₁ V₁²)/(m₂ V₂²). Using V₂ = (m₁/m₂)V₁, substitute: E₁/E₂ = (m₁ V₁²)/(m₂*(m₁²/m₂²)V₁²) )= m₂/m₁. So E₁:E₂ = m₂:m₁.
A man of weight 'W' is standing in a lift which is moving upwards with acceleration 'a'. The apparent weight of the man is
Apparent weight in an accelerating lift is W' = W(g + a)/g = W(1 + a/g).
A 4 kg mass and a 1 kg mass are moving with equal momenta. The ratio of the magnitude of their kinetic energies is
Momentum p = mv. If p is same, then KE = p²/(2m). So KE₁/KE₂ = (p²/(2m₁)) / (p²/(2m₂)) )= m₂/m₁ =1/4. So KE₁:KE₂=1:4. But the question asks for 4 kg and 1 kg. So KE4kg : KE1kg =1:4. But options are (A)4:1, (B)1:1, etc. So correct answer is A)4:1 if the question is reversed. Wait, no. If both have same momentum, the KE is inversely proportional to mass. So 4kg has KE = p²/(8), 1kg has KE = p²/(2). Ratio is (p²/8) : (p²/2) =1:4. So answer is 1:4, but this is not an option. The options are (A)4:1, (B)1:1, (C)1:2, (D)2:1. So perhaps the correct answer is A)4:1 if the question is about 1kg to 4kg. But the question says "ratio of the magnitude of their kinetic energies", so 4kg :1kg KE is 1:4. But since this is not an option, there might be a mistake. Alternatively, if the question is about 1kg :4kg, then 4:1. But the question is unclear.
A body is suspended from a rigid support by an inextensible string of length 'L' on which another identical body of mass 'm' struck inelastically moving with horizontal velocity √(2gL). The increase in the tension in the string just after it is struck by the body is
After inelastic collision, the two masses move together. Using conservation of momentum: m√(2gL) = 2m v ⇒ v=√(2gL)/2. The tension in the string provides the centripetal force: T - 2mg = 2m v²/L. Substituting v: T -2mg=2m*(2gL/4)/L=2m*(gL/2)/L= m g. So T=2mg +mg=3mg. Increase in tension is 3mg - initial tension (which was mg for single mass). But initial tension before collision was for single mass: T_initial = mg. After collision, T=3mg. So increase is 2mg. But this is not matching. Wait, the body is suspended by a string, so initial tension is mg. After collision, the combined mass is 2m moving in a circle. Tension T = 2m(g + v²/L). v=√(2gL)/2, so v²/L= (2gL)/4L= g/2. So T=2m(g + g/2)=2m*(3g/2)=3mg. Increase is 3mg - mg =2mg. But option D is 2mg. So correct answer is D)2mg.
The work done by a force on body of mass 5 kg to accelerate it in the direction of force from rest to 20 m/s² in 10 second is
Work done is force × distance. Force = m a =5 kg *20 m/s²=100 N. Distance = ½ a t²=½*20*(10)²=10*100=1000 m. Work=100*1000=10⁵ J. But this is not matching the options. Alternatively, using kinetic energy: KE=½ m v². v= a t=20*10=200 m/s. KE=½*5*(200)²=½*5*40000=100000 J=10⁵ J. But options are (A)10³, (B)4×10³, (C)2×10³, (D)10⁴. None match. Perhaps the question is about acceleration to 20 m/s (velocity), not 20 m/s². If a=20 m/s² for 10 s, v=20*10=200 m/s. But if it's acceleration to 20 m/s, then v=20 m/s. KE=½*5*(20)²=½*5*400=1000 J=10³ J. So answer A)10³ J.
A block of mass 'm' collides with another stationary block of mass '2m'. The lighter block comes to rest after collision. If the velocity of first block is 'u', then the value of coefficient of restitution is
Using conservation of momentum and coefficient of restitution. Before collision: momentum = m u. After collision: momentum =2m *v. So m u =2m v ⇒v= u/2. Coefficient of restitution e=(v -0)/(0 -u)= (u/2)/(-u)= -1/2. Taking magnitude, e=0.5.
A gardener pushes a lawn roller through a distance 20 m. If he applies a force of 30 kg-wt in a direction inclined at 60° to the ground, the work done by the gardener in pushing the roller is [g=9.8 m/s², sin30°=cos60°=½, cos30°=sin60°=√3/2]
Work done = Force × distance × cosθ. Force in newtons: 30 kg-wt =30*9.8=294 N. θ=60°, cos60°=0.5. Work=294*20*0.5=294*10=2940 J. So answer C)2940 J.
A ball at rest falls vertically on the ground from a height of 5 m. The coefficient of restitution is 0.4. The maximum height of the ball after the first rebound is [g=10 m/s²]
After collision, the velocity is e times the velocity before collision. Velocity before collision: v=√(2gh)=√(10*5)=√50≈7.07 m/s. Velocity after collision: e v=0.4*7.07≈2.83 m/s. Height after rebound: h' = v²/(2g)= (2.83)²/(2*10)=8/20=0.4 m. But this is not an option. Wait, maybe using coefficient of restitution in terms of height: h' = e² h =0.16*5=0.8 m. So answer B)0.8 m.
A ball released from a certain height strikes the ground after 2 second. After bouncing it rises to a highest point in 1 second. The coefficient of restitution is
Time to fall: 2 s. Time to rise:1 s. The velocity after bounce is e times the velocity before bounce. Time to rise is related to velocity by v =g t. So e = (v_after / v_before) = (g t_after / g t_before) = t_after / t_before =1/2=0.5.
A uniform rod of mass 'M' and length 'L' is suspended from the rigid support. A small bullet of mass 'm' hits the rod with velocity 'V' and gets embedded into the rod. The angular velocity of the system just after impact is
Using conservation of angular momentum. The bullet hits the end of the rod. Angular momentum of bullet before collision: m V * (L/2). Moment of inertia of rod about suspension point: (1/3) M L². Moment of inertia of bullet about suspension point: m (L/2)². Total moment of inertia after collision: (1/3 M L² + m (L/2)²). Angular velocity ω = (m V L/2) / ( (1/3 M L² + m L²/4 ) ) = (m V / L ) / ( (1/3 M + m/4 ) ) = (4m V / L ) / ( (4M +3m)/12 ) )= (48 m V ) / ( L (4M +3m) ) ). This doesn't match any options. Alternatively, if the rod is hit at the end, and using conservation of angular momentum: m V (L/2) = [ (1/3 M L²) + m (L/2)² ] ω. Solving for ω: ω = (m V L/2 ) / ( (1/3 M L² + m L²/4 ) ) = (m V / (2L) ) / ( (1/3 M + m/4 ) ) = (m V ) / (2L ( (4M +3m)/12 ) ) )= (6 m V ) / ( L (4M +3m) ) ). Still not matching. The options given are different. Perhaps the correct answer is C) 3mV/(M +m)L.
In a system of two particles of masses 'm₁' and 'm₂', the second particle is moved by a distance 'd' towards the centre of mass. To keep the centre of mass unchanged, the first particle will have to be moved by a distance
The center of mass must remain unchanged. If m₂ is moved towards COM by distance d, then m₁ must be moved away from COM by distance x such that m₁ x = m₂ d ⇒x= m₂ d /m₁. So first particle is moved away by x= m₂ d/m₁.
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