Question

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

Answer 1

a. 1.53
`kg(m)/(s)`

Step-by-step explanation:

According to Newton's Laws of Motion: p=mv, where p is momentum, m is mass in kg, and v is velocity in
`(m)/(s)`.

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

`v_0=0.0(m)/(s)\\a=9.8(m)/(s^2)\\\Delta y=12m\\v_f^2=(0.0(m)/(s))^2+2(9.8(m)/(s^2))(12m)\\v_f^2=235.2(m^2)/(s^2)\\v_f=\sqrt235.2(m^2)/(s^2)\\v_f=15.34(m)/(s)`

Insert this value into p=mv:

`p=(0.10kg)(15.34(m)/(s))\\p=1.53kg(m)/(s)`