Given m = mass of bullet = 50 gm = 0.50 kg M = mass of tiger = 60 kg v = Velocity of bullet = 150 m/s V = Velocity of tiger = – 10m/s (∴ It is coming from opposite direction n = no. of bullets fired per second at the tiger so as to stop it.) Pi = 0, before firing ...(i) Pf = n(mv) + MV ...(ii) ∴ FroRead more
Given m = mass of bullet = 50 gm = 0.50 kg
M = mass of tiger = 60 kg
v = Velocity of bullet = 150 m/s
V = Velocity of tiger = – 10m/s
(∴ It is coming from opposite direction n = no. of bullets fired per second
at the tiger so as to stop it.)
Pi = 0, before firing …(i)
Pf = n(mv) + MV …(ii)
∴ From the law of conservation of momentum,
Pi = Pf
⇒ 0 = n(mv) + MV
⇒ n = MV/mv
= -60/(-10)/0.05×150 = 80.
(a) Force on the floor of the helicopter by the crew and passengers = apparent weight of crew and passengers = 500 (10 + 15) = 12500 N (b) Action of rotor of helicopter on surrounding air is Obviously vertically downwards, because helicopter rises on account of reaction of this force. Thus force ofRead more
(a) Force on the floor of the helicopter by the crew and passengers
= apparent weight of crew and passengers
= 500 (10 + 15)
= 12500 N
(b) Action of rotor of helicopter on surrounding air is Obviously vertically
downwards, because helicopter rises on account of reaction of this
force. Thus force of action
= (2000 + 500) (10 + 15)
= 2500 × 25
= 62,500 N
(c) Force on the helicopter due to surrounding air is obviously a reaction.
As action and reaction are equal and opposite, therefore
Force of reaction F´ = 62,500 vertically upwards.
Suppose, a = acceleration produced if m₁ and m₂ are tied together, F = 100 N Let a₁ and a₂ be the acceleration produced in m₁ and m₂, respectively. a₁ = 10 ms⁻², a₂ = 20 ms⁻² (given) Again m₁ = F/a₁ and m₂ = F/ a₂ m₁ = 100/10 = 10 ms⁻² and m₂ = 100/20 = 5 ms⁻² m₁ + m₂ = 10 + 5 = 15 So, a = F/m₁ + m₂Read more
Suppose, a = acceleration produced if m₁ and m₂ are tied together,
F = 100 N
Let a₁ and a₂ be the acceleration produced in m₁ and m₂, respectively.
a₁ = 10 ms⁻², a₂ = 20 ms⁻² (given)
Again m₁ = F/a₁ and m₂ = F/ a₂
m₁ = 100/10 = 10 ms⁻²
and m₂ = 100/20 = 5 ms⁻²
m₁ + m₂ = 10 + 5 = 15
So, a = F/m₁ + m₂ = 100/15 = 20/3
= 6.67 ms⁻²
Force of friction = 0·5 N per quintal f = 0·5 × 2000 = 1000 N m = 2000 quintals = 2000 × 100 kg sin θ = 1/50, a= 2m/s² In moving up an inclined plane, force required against gravity = mg sin θ = 39200 N And force required to produce acceleration = ma = 2000 × 100 × 2 = 40,0000 N Total force requiredRead more
Force of friction = 0·5 N per quintal
f = 0·5 × 2000 = 1000 N
m = 2000 quintals = 2000 × 100 kg
sin θ = 1/50, a= 2m/s²
In moving up an inclined plane, force required against gravity
= mg sin θ = 39200 N
And force required to produce acceleration = ma
= 2000 × 100 × 2 = 40,0000 N
Total force required = 1000 + 39,200 + 40,0000
= 440200 N.
A hunter has a maching gun that can fire 50g bullets with a velocity of 150 ms⁻¹. A 60 kg tiger springs at him with a velocity of 10 ms⁻¹. How many bullets must the hunter fire into the target so as to stop him in his track ?
Given m = mass of bullet = 50 gm = 0.50 kg M = mass of tiger = 60 kg v = Velocity of bullet = 150 m/s V = Velocity of tiger = – 10m/s (∴ It is coming from opposite direction n = no. of bullets fired per second at the tiger so as to stop it.) Pi = 0, before firing ...(i) Pf = n(mv) + MV ...(ii) ∴ FroRead more
Given m = mass of bullet = 50 gm = 0.50 kg
See lessM = mass of tiger = 60 kg
v = Velocity of bullet = 150 m/s
V = Velocity of tiger = – 10m/s
(∴ It is coming from opposite direction n = no. of bullets fired per second
at the tiger so as to stop it.)
Pi = 0, before firing …(i)
Pf = n(mv) + MV …(ii)
∴ From the law of conservation of momentum,
Pi = Pf
⇒ 0 = n(mv) + MV
⇒ n = MV/mv
= -60/(-10)/0.05×150 = 80.
A helicopter of mass 2000 kg rises with a vertical acceleration of 15 m/s². The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the : (i) Force on the floor of the helicopter by the crew and passenger. (ii) Action of the rotor of the helicopter on the surrounding air (iii) Force on the helicopter due to the surrounding air (g =10 m/s²)
(a) Force on the floor of the helicopter by the crew and passengers = apparent weight of crew and passengers = 500 (10 + 15) = 12500 N (b) Action of rotor of helicopter on surrounding air is Obviously vertically downwards, because helicopter rises on account of reaction of this force. Thus force ofRead more
(a) Force on the floor of the helicopter by the crew and passengers
See less= apparent weight of crew and passengers
= 500 (10 + 15)
= 12500 N
(b) Action of rotor of helicopter on surrounding air is Obviously vertically
downwards, because helicopter rises on account of reaction of this
force. Thus force of action
= (2000 + 500) (10 + 15)
= 2500 × 25
= 62,500 N
(c) Force on the helicopter due to surrounding air is obviously a reaction.
As action and reaction are equal and opposite, therefore
Force of reaction F´ = 62,500 vertically upwards.
A force of 100 N gives a mass m1, an acceleration of 10 ms⁻² and of 20 ms⁻² to a mass m₂. What acceleration must be given to it if both the masses are tied together?
Suppose, a = acceleration produced if m₁ and m₂ are tied together, F = 100 N Let a₁ and a₂ be the acceleration produced in m₁ and m₂, respectively. a₁ = 10 ms⁻², a₂ = 20 ms⁻² (given) Again m₁ = F/a₁ and m₂ = F/ a₂ m₁ = 100/10 = 10 ms⁻² and m₂ = 100/20 = 5 ms⁻² m₁ + m₂ = 10 + 5 = 15 So, a = F/m₁ + m₂Read more
Suppose, a = acceleration produced if m₁ and m₂ are tied together,
See lessF = 100 N
Let a₁ and a₂ be the acceleration produced in m₁ and m₂, respectively.
a₁ = 10 ms⁻², a₂ = 20 ms⁻² (given)
Again m₁ = F/a₁ and m₂ = F/ a₂
m₁ = 100/10 = 10 ms⁻²
and m₂ = 100/20 = 5 ms⁻²
m₁ + m₂ = 10 + 5 = 15
So, a = F/m₁ + m₂ = 100/15 = 20/3
= 6.67 ms⁻²
Calculate the force required to move a train of 2000 quintal up on an incline plane of 1 in 50 with an acceleration of 2 ms⁻². The force of friction per quintal is 0.5 N.
Force of friction = 0·5 N per quintal f = 0·5 × 2000 = 1000 N m = 2000 quintals = 2000 × 100 kg sin θ = 1/50, a= 2m/s² In moving up an inclined plane, force required against gravity = mg sin θ = 39200 N And force required to produce acceleration = ma = 2000 × 100 × 2 = 40,0000 N Total force requiredRead more
Force of friction = 0·5 N per quintal
See lessf = 0·5 × 2000 = 1000 N
m = 2000 quintals = 2000 × 100 kg
sin θ = 1/50, a= 2m/s²
In moving up an inclined plane, force required against gravity
= mg sin θ = 39200 N
And force required to produce acceleration = ma
= 2000 × 100 × 2 = 40,0000 N
Total force required = 1000 + 39,200 + 40,0000
= 440200 N.
A force of 98 N is just required to move a mass of 45 kg on a rough horizontal surface. Find the coefficient of friction and angle of friction?
F = 98 N, R = 45 × 9·8 = 441 N μ² = F'/R = 0.22 Angle of friction θ = tan⁻¹ μ = tan⁻¹ 0·22 = 12°24′
F = 98 N, R = 45 × 9·8 = 441 N
See lessμ² = F’/R = 0.22
Angle of friction θ = tan⁻¹ μ = tan⁻¹ 0·22 = 12°24′