Question

A stationary 1.67-kg object is struck by a stick. The object experiences a horizontal force given by F = at - bt2, where t is the time in milliseconds from the instant the stick first contacts the object.
If :

a = 1500 N/(ms)

b = 20 N/(ms)2

what is the speed of the object just after it comes away from the stick at t = 2.74 ms?

Answer 1

Answer:

$v_{f}$  = 3289.8 m / s

Explanation:

This exercise can be solved using the definition of momentum

     I = ∫ F dt

Let's replace and calculate

     I = ∫ (at - bt²) dt

We integrate

      I = a t² / 2 - b t³ / 3

We evaluate between the lower limits I=0  for t = 0 s and higher I=I for t = 2.74 ms

      I = a (2,74² / 2- 0) - b (2,74³ / 3 -0)

      I = a 3,754 - b 6,857

We substitute the values ​​of a and b

      I = 1500 3,754 - 20 6,857

      I = 5,631 - 137.14

      I = 5493.9 N s

Now let's use the relationship between momentum and momentum

      I = Δp = m $v_{f}$ - m v₀o

      I = m $v_{f}$  - 0

     $v_{f}$  = I / m

    $v_{f}$  = 5493.9 /1.67

    $v_{f}$  = 3289.8 m / s