angel asked in Science & MathematicsPhysics · 3 months ago

Physics HW Help ?

A crane drops a 0.3 kg steel ball onto a steel plate. The ball’s speed just before impact

and after impact is -4.5 m/s and 4.2 m/s respectively. If the ball is in contact with the

plate for 0.03 s, what is the magnitude of the average force that the ball exerts on the

plate during impact?

a. 87 N

b. 133 N

c. 3.0 N

d. 3.5 N

I dont understand how to find magnitude.

Usually isn't it square root of x +y ? but here i dont know how that applies

3 Answers

  • NCS
    Lv 7
    3 months ago

    There are two expressions for "impulse" (change in momentum) --

    Δp = mΔv


    Δp = FΔt

    Equate them and plug in your data and you get

    0.3kg * (4.2 - -4.5)m/s = F * 0.03s

    F = 87 N

    √(x²+y²) doesn't come into play because you have no "x" component in this problem. The velocity changes from the -y direction to the +y direction.

  • Amy
    Lv 7
    3 months ago

    Magnitude just means size. In two dimensions you have to combine x and y to get the length of the diagonal. In one dimension it just means you don't have to describe the direction of the force.

  • oubaas
    Lv 7
    3 months ago

    Change in momentum ΔM = 0.3*(4.2-(-4.5)) =2.61 kg*m/sec = F*Δt

    F = Δ M / Δ t = 2.61*100/3 = 87 N 

Still have questions? Get answers by asking now.