Camera Dynamics
Dynamics qn??? How do you find Impulsive Force (in Newtons)?
A mobile phone is dropped to a hard floor from a height of h m, with h as given below. The mass of the mobile phone is 200 grams. After the impact, the phone jumps back by 5 cm. The impact is captured by a high-speed camera and the duration of the time while the phone was in contact with the floor is estimated as 0.005 s. What is the average value of the impulsive force (in N)?
h[m] = 0.7;
speed of phone as it strikes floor = √2gh = √19.6(0.7) = 3.70 m/s
speed of phone as it leaves floor = √2gh = √19.6(0.05) = 0.99 m/s
m1v1 = (0.200)(3.70) = 0.741 kg-m/s
m2v2 = (0.200)(0.99) = 0.198 kg-m/s
m1v1 – m2v2 = 0.741 – 0.198 = 0.543
Impulse = Favg (t) = m1v1 – m2v2
0.005Favg = 0.543
Favg = 0.543/0.005 = 109 N ANS
Camera Dynamics at Roscor2010