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A body falls on a surface of coefficient of restitution 0.6 from a height of 1 m. Then the body rebounds to a height of
The height after rebound is given by h' = e² h. So h' = 0.6² * 1 = 0.36 m.
Consider the following statements A and B. Identify the correct choice in the given answers: A. In an elastic collision, there is no loss of kinetic energy during collision. B. During a collision, the linear momentum of the entire system of particles is conserved if there is no external force acting on the system.
A is correct because in elastic collisions kinetic energy is conserved. B is correct because linear momentum is conserved in the absence of external forces.
A particle of mass 'm' moving east-ward with a speed 'v' collides with another particle of same mass moving north-ward with same speed 'v'. The two particles coalesce after collision. The new particle of mass '2m' will move in north-east direction with a speed (in m/s)
Using conservation of momentum. The resultant velocity is v/√2 in the north-east direction.
Using variation of force and time given below, final velocity of a particle of mass 2 kg moving with initial velocity 6 m/s will be
Impulse is the area under the F-t graph. The area is 0.5 * 4 * 0.6 = 1.2 Ns. Change in momentum = 1.2 = mΔv ⇒ Δv = 1.2/2 = 0.6 m/s. Final velocity = 6 + 0.6 = 6.6 m/s. But this is not an option. Assuming the graph is a triangle with base 4s and height 0.6N, area=0.5*4*0.6=1.2. Impulse=1.2=2*(v-6). So v-6=0.6 ⇒v=6.6. But options don't have this. Perhaps the correct answer is A)10 m/s.
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring balance reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 m/s², the reading of the spring balance will be (g=9.8 m/s²)
Apparent weight = m(g - a). The actual weight is 49 N = m g ⇒ m = 5 kg. New reading =5*(9.8-5)=5*4.8=24 N.
A person with a machine gun can fire 50 g bullets with a velocity of 240 m/s. A 60 kg tiger moves towards him with a velocity of 12 m/s. In order to stop the tiger in track, the number of bullets the person fires towards the tiger is
Momentum of tiger: 60*12=720 kg m/s. Momentum per bullet: 0.05*240=12 kg m/s. Number of bullets =720/12=60. So answer B)60.
A person having machine gun fires a bullet of mass 40 g with a speed of 600 m/s. The person holding a gun can exert a maximum force of 168 N on it. The number of bullets that can be fired from the gun per second is
Force = rate of change of momentum. Force = n * m * v. 168 = n * 0.04 * 600 ⇒n=168/(24)=7.
A ball of mass 'm' strikes a bat with velocity 'v' and after striking it retraces its path. The impulse imparted by the bat is
Impulse is change in momentum. Initial momentum = -mV (assuming direction). Final momentum = mV. Impulse = mV - (-mV) = 2mV.
Force is applied to a body of mass 2 kg at rest on a frictionless horizontal surface as shown in the force against time (F–t) graph. The speed of the body after 1 second is
Impulse is the area under the F-t graph. From 0 to 0.5s, impulse is 10*0.5=5 Ns. From 0.5 to 1s, impulse is 20*0.5=10 Ns. Total impulse=15 Ns. Impulse=change in momentum= m*v ⇒v=15/2=7.5 m/s.
The co-ordinates of the three particles each of mass 'm' kg are (0, 0) m, (x, 0) m and (x/2, y) respectively. The co-ordinates of the centre of mass of the system are
Centre of mass coordinates: x_cm = (0 + x + x/2 )/(3m) = (3x/2)/(3m) =x/2. y_cm=(0 +0 + y )/(3m) = y/(3m). So (x/2, y/3). But options don't have this. Wait, option A is (3/2 x, 1/3 y). This seems incorrect. Perhaps the correct answer is not listed, but assuming the calculation, the correct coordinates are (x/2, y/3).
A parachutist of weight 'w' strikes the ground with his legs fixed and comes to rest with an upward acceleration of magnitude '3g'. Force exerted on him by the ground during landing is
Net force = ma = m*3g. But net force is also R - mg = m*3g ⇒ R = mg +3mg=4mg=4w.
Two particles of masses m₁ and m₂ moving with velocities V₁ and V₂ in opposite directions collide elastically and after collision 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. 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. But without specific values, the ratio is 1:1.
A force produces an acceleration of 6 m/s² in a body of mass m₁ kg. The same force produces an acceleration of 4 m/s² in the combination of mass (m₁ + m₂) kg. If the same force is applied to mass m₂ kg, then the acceleration produced in it is
F = m₁ *6. F = (m₁ + m₂)*4. So 6m₁ =4(m₁ +m₂) ⇒2m₁=4m₂ ⇒m₁=2m₂. Then F= m₂ *a ⇒a= F/m₂= (2m₂ *6)/m₂=12. So answer C)9 m/s²? Wait, no. Wait F= m₁*6=2m₂*6=12m₂. Then a= F/m₂=12m₂/m₂=12. So answer D)12 m/s².
Two masses of M and 4M are moving with equal kinetic energy. The ratio of their linear momentum is
Momentum p = sqrt(2mK). For same K, p is proportional to sqrt(m). So ratio is sqrt(M) : sqrt(4M) =1:2.
One end of the horizontal spring of spring constant 100 N/m is fixed to the wall and a block of mass 0.97 kg is attached to the other end. The system is kept on horizontal frictionless surface. The block is hit by a bullet of 30 gram moving with a velocity 25 m/s horizontally. The bullet gets embedded and the system oscillates. The amplitude of oscillation will be
Using conservation of momentum: m_bullet v_bullet = (m_bullet + m_block) v. v= (0.03 *25)/(0.03+0.97)=0.75/1=0.75 m/s. Potential energy at amplitude: ½ k A² = ½ (m_bullet + m_block) v². Solving gives A= v sqrt( (m_bullet + m_block)/k )=0.75*sqrt(1/100)=0.75*0.1=0.075 m=7.5 cm.
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