A ball with a mass of 0.10 kg is dropped from a height of 12 m. What is the magnitude of its momentum when it strikes the ground? (Hint: The ball is in free-fall: Use the appropriate kinematical equation )
a. 1.5 kg-m/s
b. 1.8 kg-m/s
c. 2.4 kg-m/s
d. 4.8 kg-m/s

Respuesta :

Answer:

a. 1.53 [tex]kg\frac{m}{s}[/tex]

Explanation:

According to Newton's Laws of Motion: p=mv, where p is momentum, m is mass in kg, and v is velocity in [tex]\frac{m}{s}[/tex].

To find the velocity, use kinematic equation #4: [tex]v_f^2=v_0^2+2a\Delta y[/tex]

[tex]v_0=0.0\frac{m}{s}\\a=9.8\frac{m}{s^2}\\\Delta y=12m\\v_f^2=(0.0\frac{m}{s})^2+2(9.8\frac{m}{s^2})(12m)\\v_f^2=235.2\frac{m^2}{s^2}\\v_f=\sqrt235.2\frac{m^2}{s^2}\\v_f=15.34\frac{m}{s}[/tex]

Insert this value into p=mv:

[tex]p=(0.10kg)(15.34\frac{m}{s})\\p=1.53kg\frac{m}{s}[/tex]