A stone is dropped from a height of h_1 = 335 m above the ground. After it bounc

Question

A stone is dropped from a height of h_1 = 335 m above the ground. After it bounces, it only makes it to a height h_2=22 m above .the ground. The stone has mass m = 0.116 kg. Randomized Variables h_1 =335 m h_2 - 22 m m= 0.116 kg What is the magnitude of the impulse l, in kilogram meters per second, the stone experienced during the bounce? If the stone was in contact with the ground for t = 0J61 s, what was the magnitude of the constant force F acting on it, in newtons? How much energy, in joules, did the stone transfer to the environment during the bounce?

Explanation / Answer

problem 1:

initial speed before hitting the ground=sqrt(2*g*h1)=sqrt(2*9.8*3.35)=8.1031 m/s

speed after bouncing from the ground=sqrt(2*g*h2)=sqrt(2*9.8*2.2)=6.5665 m/s

(as the direction of the motion after bouncing is oppoiste to initial direction of motion before hitting the ground,

the veloicty of speed after bounce=-6.5665 m/s)

then impulse applied on the stone=change in its momentum=mass*change in veloicty

=0.116*(8.1031+6.5665)=1.7017 kg.m/s

problem 2:

time of contact=0.161 seconds

hence force applied=impulse/time of contact=1.057 N

problem 3:

energy transferred=change in kinetic energy of the stone

=0.5*mass*(initial speed^2-final speed^2)=0.5*0.116*(8.1031^2-6.5665^2)=1.3074 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.