A body is projected at an angle θ to the horizontal with momentum of magnitude p. The magnitude of momentum a?
A body is projected at an angle θ to the horizontal with momentum of magnitude p. The magnitude of momentum at the topmost point will be………………………?
More Questions From This User See All
Answers & Comments
Verified answer
Momentum = mass * velocity
The initial velocity of the body is p ÷ mass
As the body moves upward to the topmost point, the vertical component of the velocity decreases 9.8 m/s each second. When the body reaches the topmost point, the vertical component of the velocity is 0 m/s.
The momentum at the topmost point = mass * horizontal component of the velocity =
The horizontal component of the velocity remains constant for the entire trip.
The horizontal component of the velocity = Initial velocity * cos θ
Initial velocity = p ÷ mass
The horizontal component of the velocity = (p ÷ mass) * cos θ
The momentum at the topmost point = mass * (p ÷ mass) * cos θ = p * cos θ
Due to the fact that momentum is a vector = m x pace vector = m (xi + yj) consequently p(vector) = mxi + myj and p(magnitude) = sqrt[ (mx)2^+ (my)^2] = m sqrt(x^2 + y^2) =0.5456sqrt(three.886^2 + 4.43^2)= 3.22 Ns