Question

1. A 21.3-kg child positions himself on an inner-tube which is suspended by a 7.28-m long rope attached to a strong tree limb. The child and tube is drawn back until it makes a 17.4-degree angle with the vertical. The child is released and allowed to swing to and from. Assuming negligible friction, determine the child's speed at his lowest point in the trajectory.

2. A baseball player catches a 163-gram baseball which is moving horizontally at a speed of 39.8 m/s. Determine the force which she must apply to the baseball if her mitt recoils a horizontal distance of 25.1 cm.

Answer 1

Answer

1) given,

mass of child = 21.3 Kg

length of the rope = L = 7.28 m

angle made with vertical = 17.4°

speed of the child at the bottom most point

using conservation energy

$m g h = \dfrac{1}{2}mv^2$

$v = \sqrt{2gh}$

H = L cos θ

H = 7.28 x cos 17.4°

H = 6.94

where

L = H + h

h = L - H

h = 7.28 - 6.94 = 0.34 m

now,

$v = \sqrt{2\times 9.8 \times 0.34}$

v = 2.58 m/s

2) given,

   mass of baseball = 163 g = 0.163 kg

  initial speed = 39.8 m/s

  final speed = 0

  horizontal distance = 25.1 cm = 0.251 m

   Force = ?

 using equation of motion

     v = u + at

     0 = 39.8 + at

     at = -39.8 m/s

 using equation

  $s = ut + \dfrac{1}{2}at^2$

  $0.251= 39.8 t - 0.5 \times 39.8 t$

     t = 0.0126 s

using formula of impulse

 I = F x t

 I = m(v - u)

F x 0.0126 = 0.163 x (-39.8)

 F = 514.87 N