Solving Mass from 2 Forces & Force of Impacting Bullet

[omonoid]

Part 1

1. The only two forces acting on a body have magnitudes of 20 N and 42 N and directions that differ by 60°. The resulting acceleration has a magnitude of 20 m/s2. What is the mass of the body?2. F=ma3. I am not sure how to solve this because I'm used to being able to use trig like cos and tan, but this problem doesn't create a triangle with 90 degrees

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Part 2

1. A bullet of mass 6.5*10-3 kg moving at 1150 m/s impacts with a large fixed block of wood and travels 8 cm before coming to rest. Assuming that the deceleration of the bullet is constant, find the force exerted by the wood on the bullet. 2. F=ma
V^2=Vi^2+2ax3. V=0 Vi=1150 m/s x=.08 m solve for a and i got -8265625 m/s^2 then multiply by mass (6.5E-3) and get -53726.5625. Is this right? would i void the negative on a force?

 

[alphysicist]

Hi omonoid,

Part 1

1. The only two forces acting on a body have magnitudes of 20 N and 42 N and directions that differ by 60°. The resulting acceleration has a magnitude of 20 m/s2. What is the mass of the body?


2. F=ma

Let me be picky here and say you should write this as F_net=ma, that is, the sum of the forces equals m*a. Does that help? Have you added vectors together that were not perpendicular to each other?

3. I am not sure how to solve this because I'm used to being able to use trig like cos and tan, but this problem doesn't create a triangle with 90 degrees

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Part 2

1. A bullet of mass 6.5*10-3 kg moving at 1150 m/s impacts with a large fixed block of wood and travels 8 cm before coming to rest. Assuming that the deceleration of the bullet is constant, find the force exerted by the wood on the bullet.


2. F=ma
V^2=Vi^2+2ax


3. V=0 Vi=1150 m/s x=.08 m solve for a and i got -8265625 m/s^2 then multiply by mass (6.5E-3) and get -53726.5625. Is this right? would i void the negative on a force?

That number looks right to me. The minus sign is showing that the force is in the opposite direction of the initial velocity.