Ok so, I have two questions. I have already tried and done my research, but I ju

Question

Ok so, I have two questions. I have already tried and done my research, but I just need it "dummy-proofed" lol. I know these have been asked on here before, but I didn't see what I needed answered so I can understand better.

1) A solid sphere of mass 4.0 kg and radius of .12 m is at rest at the top of a ramp inclined 15degrees. It rolls to the bottom without slipping. The upper end of the ramp is 1.2 m higher than the lower end. What is the linear speed of the sphere when it reaches the bottom of the ramp?

The answer is 4.1 m/s.... so here is what i have so far.

I have 9.8(1.2)=11.76

11.76/.7 = 16.8

sqrt 16.8 = 4.1

Now, HOW do you get the .7? From the .7 I can figure it out. Please help me understand this.

2) A 25.0 g bullet is moving at 900 m/s is fired through a .900 kg block of wood emerging at a speed of 100 m/s. If the block had been originally at rest and is free to move, what is it's resulting speed?

Answer is 22 m/s.

I have NO idea where to even start. Any help would be great.

Explanation / Answer

there is another way to solve

in case rolling witout slipping

rotational energy/translational energy=xmr^2 (where x is 2/5 in this case,bacuase MI of solid sphere is 2/5 mr^2)

than total energy at bottom=E rotational+E translational=2/5 E transla+E transl=7/5 translational=7/5*.5mv^2=.7mv^2

aplly energy rule

mgh=.7mv^2

9.8*1.2=.7v^2

v=4.0987

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

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