Question

22)A child bounces a 60-gram superball on the sidewalk. The velocity change of the superball is from 20 m/s downward to 18 m/s upward. If the contact time with the side walk is 1/800 S, what is the magnitude of the force exerted on the superball by the sidewalk? a) 96 N b) 1824 N c) 106 N d) 1211 N 23)Three particles (point masses) of mass 2 kg, 3 kg, and 5 kg are welded to a straight massless rod as shown in the figure. Find the location of the center of mass of the assembly. a) 0.26 m b) 0.16 m 2m 2kg 3 kg 5kg c) 0.2 m d) 0.13 24)What is the mechanical advantage of a nail puller-similar to the one shown in the figure below-where you exert a force 30 cm from the pivot and the nail is 1.1 cm on the other side? a) 0.036 b) 33 c) 31.1 d) 27.3 Freebody dagram 25)Suppose that a meter stick is supported at the center, and a 20-N block is hung at the 80- cm mark. Another block of unknown weight just balances the system when it is hung at the 10-cm mark. What is the weight of the second block? a) 160 N b) 400 N c) 15 N d) 30 N 60 80 40 100 ?N 20 N

Answer 1

Question 22)

Given:-

• Mass of the superball :
  m = 60 gram = 0.06 kg
  
• Downward velocity       : $\small v_{d} = 20 \; m/s$
• Upward velocity           : $\small v_{u} = 18 \; m/s$
• Contact time               : $\small \Delta t = 1/800 \; s$

We will consider the upward direction as positive and downward as negative. Therefore, the downward velocity will be negative,

Vo = -20 m/s

Impulse is given by,

$i = F \times \Delta t = \Delta P$

$F \times \Delta t = \Delta P$

Where,

• $\small F$ : is the force exerted on the ball.
• $\small \Delta P$ : is the change in momentum.

$F \times \Delta t = m v_{u} - m v _{d}$

$F = \frac{m \left (v_{u} - v _{d} \right )}{ \Delta t}$

$F = \frac{0.06 \left (18 - \left ( -20 \right ) \right )}{ 1/800}$

$F = 0.06 \left (38 \right ) \times 800$

$\mathbf{F = 1824 \; N}$

This is the magnitude of the force exerted on the superball by the sidewalk.

Question 23)

Given:-

• Mass of first particle : $\small m_{1} = 2 \; kg$
• Mass of second particle : $\small m_{2} = 3 \; kg$
• Mass of third particle : $\small m_{3} = 5 \; kg$
• Distance of first particle from the left : $\small x_{1} = 0$
• Distance of second particle from the left : $\small x_{2} = 0.2 \; m$
• Distance of third particle from the left : $\small x_{3} = 0.4 \; m$

The location of the center of mass is given by,

$x_{cm} = \frac{m _{1} x_{1} + m _{2} x_{2} +m _{3} x_{3}}{m _{total}}$

$x_{cm} = \frac{m _{1} x_{1} + m _{2} x_{2} +m _{3} x_{3}}{m _{1} +m _{2} + m _{3}}$

$x_{cm} = \frac{\left (2 \times 0 \right ) + \left (3 \times 0.2 \right ) +\left (5 \times 0.4 \right )}{2+3+5}$

$x_{cm} = \frac{2.6}{10}$

$\mathbf{x_{cm} =0.26 \; m}$

This is the location of the center of mass of the assembly from the left.

Question 24)

Given:-

• Distance from where input force applied : $\small L_{i}= 30 \; cm$
• Distance from where output force applied : $\small L_{o}= 1.1 \; cm$

Mechanical advantage is given by,

$MA = \frac{L_{i}}{L_{o}}$

$MA = \frac{30 }{1.1}$

$\mathbf{MA = 27.3 }$

This is the mechanical advantage of a nail puller.

Question 25)

Given:-

• Weight of the block 1 : $\small W_{1} = 20 \; N$
• Distance of this block 1 from the center of mass : $\small r_{1} = 30 \; cm = 0.3 \; m$
• Distance of this block 2 from the center of mass : $\small r_{2} = 40 \; cm = 0.4 \; m$

Let, $\small W_{2}$ be the unknown weight of the block 2.

Since the meter stick is in balance, the total torque will be zero.

$\sum \tau = 0$

$\tau _{1} + \left (- \tau _{2} \right )= 0$

$W_{1} \times r_{1} - W_{2} \times r_{2}= 0$

$W_{2} = \frac{W_{1} \times r_{1} }{r_{2}}$

$W_{2} = \frac{20 \times 0.3 }{0.4}$

$\mathbf{W_{2} = 15 \; N}$

This is the weight of the second block.