Physics-
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Question

When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , Mass of cart M = 10 kg: Velocity of cart at t = 10 s. is equal to:

  1. 4 m/s    
  2. 6 m/s    
  3. 8 m/s    
  4. 5 m/s    

The correct answer is: 8 m/s

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When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , Mass of cart M = 10 kg: Initially (t = 0) the force on the cart is equal to :

When jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F = open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript minus open parentheses rho A v subscript 0 end subscript close parentheses v subscript 0 end subscript c o s invisible function application theta equals rho A v subscript 0 end subscript superscript 2 end superscript left square bracket 1 minus c o s invisible function application theta right square bracket

If surface is free and starts moving due to thrust of liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let any instant the velocity of surface is u, then above equation becomes – F equals rho A open parentheses v subscript 0 end subscript minus u close parentheses to the power of 2 end exponent left square bracket 1 minus c o s invisible function application theta right square bracket

Based on above concept, in the below given figure, if the cart is frictionless and free to move in horizontal direction, then answer the following :Given cross–section area of jet equals 2 cross times 10 to the power of negative 4 end exponent m to the power of 2 end exponent velocity of jet V subscript 0 end subscript = 10 m/s, density of liquid = 1000 kg/ m to the power of 3 end exponent , Mass of cart M = 10 kg: Initially (t = 0) the force on the cart is equal to :

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