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Grade 12Mechanics

a body of mass 0.5 kg is project under gravity with a speed of 98 meter per second at an angle of 30 degree with the horizontal. the change in momentum of body is

Profile image of Shashidhar k
9 Years agoGrade 12
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1 Answer

Profile image of Vikas TU
9 Years ago
Time of flight = 2usinthetha/g => 2*98*sin30/g => 9.8 seconds.
Now in x – direction final velcity would be =>
Vx = 98cos30 => 98*root(3)/2 => 49root(3)
Vy = 98sin30 – g*9.8 => 49 – 96.04 => -47 m/s (in the negative y direction)
Vfinalnet = root(7203 + 2209) => root(9412) = 97 m/s
 
Hence the change in momentum is => 0.5*(vf – vi) => 0.5(97 – 98) => -0.5 kg*m/s