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PHY 111 Final Exam Questions And Answers 100% Verified. A .toy .car .with .mass .m1 .(0.36 .kg .) .travels .to .the .right .on .a .frictionless .track .with .a .speed .of .3 .m/s. .A .second .toy .car .with .a .mass .m2 .(0.74 .kg) .travels .with .a .speed .of .6 .m/s .to .the .left .on .t...

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  • August 23, 2024
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  • phy 111 final exam questions and answers 100 ver
  • a toy car with mass m1 036 kg travels to the r
  • a student fires a 007 kg arrow at an object with

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PHY 111 Final Exam Questions And
Answers 100% Verified.

A .toy .car .with .mass .m1 .(0.36 .kg .) .travels .to .the .right .on .a .frictionless .track .with .a .speed .of
.3 .m/s. .A .second .toy .car .with .a .mass .m2 .(0.74 .kg) .travels .with .a .speed .of .6 .m/s .to .the .left

.on .the .same .track. .The .two .cars .collide .and .stick .together. .What .are .the .cars' .speeds

.after .the .collision? .- .correct .answer. . . . .momentum .conservation .eq:

m1v1 .- .m2v2 .= .(m1 .+ .m2)v

where: .m1 .= .0.36 .kg, .v1 .= .3 .m/s, .m2 .= .0.74 .kig, .v2 .= .6 .m/s

(0.36)(3) .- .(0.74)(6) .= .(0.36 .+ .0.74)v
v .= .3.05 .m/s

A .student .fires .a .0.07 .kg .arrow .at .an .object .with .mass .m .(6.1 .kg) .that .is .initially .at .rest .on
.a .frictionless .surface. .The .speed .of .the .arrow .before .the .collision .is .90 .m/s. .The .speed

.when .the .arrow .emerges .from .the .object .is .v .(0.78 .m/s). .What .is .the .resulting .velocity .of

.the .object?




b) .Is .the .collision .between .the .arrow .and .the .object .elastic .or .inelastic? .Include .evidence
.to .support .your .answer. .- .correct .answer. . . . .eq: .m1v1 .+ .m2v2 .= .(m1)(v1A) .+ .(m2)(V2A)




where: .m1 .= .0.7 .kg, .v1 .= .90 .m/s, .m2 .= .6.1 .kg, .v1A .= .0.78 .m/s

(0.7)(90) .+ .0 .= .(0.7)(.78) .+ .(6.1)(v2A)
v2A .= .10.24 .m/s

b) .The .initial .ad .final .kinetic .energy .are .not .equal .(calculated .using .the .following .eq:
.1/2(m)(v)^2). .This .means .that .the .collision .is .inelastic .as .it .lost .some .kinetic .energy

.through .the .collision.

,A .class .F .model .rocket .engine .can .provide .an .impulse .between .40-80 .N-s. .A .student
.attaches .a .class .F .rocket .that .can .apply .an .impulse .of .60 .N-s .to .a .model .rocket .of .mass

.(m .= .2 .kg) .If .the .engine .burns .for .(t .= .7 .seconds), .what .is .the .average .engine .thrust

.provided .by .the .engine?




b) .What .maximum .speed .can .the .rocket .reach? .(Assume .that .air .resistance .is .negligible.)
.- .correct .answer. . . . .eq: .J .(impulse) .= .(F)(t)




where: .impulse .= .60 .N-s, .t .= .7 .seconds

(60) .= .F(7)
F .= .8.57 .N

b) .eq: .J .(impulse) .= .m(delta .v)

where: .impulse .= .60 .N-s, .m .= .2 .kg

(60) .= .(2)(v)
v .= .30 .m/s

Three .particles .are .held .at .rest .at .position .(0,0) .m. .When .released, .the .particles .apply .a
.momentary .repelling .force .on .each .other. .Particle .A .has .a .mass .of .(mA .= .1.2 .kg) .and .has

.a .final .velocity .of .(vA .= .2.0 .m/s) .at .70*. .Particle .B .has .a .mass .of .(mB .= .1.6 .kg)and .has .a

.final .velocity .of .(vB .= .3.1 .m/s) .at .270*. .Particle .C .has .a .mass .of .(mC .= .2.6 .kg). .What .are

.the .horizontal .and .vertical .components .of .particle .C's .final .velocity? .




b) .What .is .the .velocity .of .particle .C .(magnitude .and .direction)? .- .correct .answer. . . .
.horizontal .eq: .mv .+ .0 .= .(m .+ .m)vx




where: .m .= .1.2 .kg, .v .= .2 .m/s, .m2 .= .1.6 .kg

(1.2)(2) .+ .0 .= .(1.2 .+ .1.6)vx
vx .= .0.86 .m/s

vertical .eq: .0 .+ .mv .= .(m .+ .m)vy

where: .m .= .1.6 .kg, .v .= .3.1 .m/s, .m2 .= .1.2 .kg

0 .+ .(1.6)(3.1) .= .(1.6 .+ .1.2)vy
vy .= .1.77 .m/s

b) .Pythagorean .theorem .eq: .a^2 .+ .b^2 .= .c^2

where: .a .= .0.86 .m/s, .b .= .1.77 .m/s

(0.86)^2 .+ .(1.77)^2 .= .v^2

, v .= .1.97 .m/s
tan .(1.77/0.86) .= .64. .11 .degrees

An .inattentive .car .driver .crashes .into .their .neighbor's .mailbox. .Compare .the .force .of .the
.car .on .the .mailbox .to .the .force .of .the .mailbox .on .the .car. .Explain .your .reasoning. .




b) .Compare .the .change .in .momentum .experienced .by .the .car .to .the .change .in
.momentum .experienced .by .the .mailbox. .Explain .your .reasoning. .




c) .Compare .the .change .in .velocity .experienced .by .the .car .to .the .change .in .velocity
.experienced .by .the .mailbox. .Explain .your .reasoning. .- .correct .answer. . . . .The .force .of .the

.car .on .the .mailbox .will .be .equal .to .the .force .of .the .mailbox .on .the .car .(when .the .car .hits

.the .mailbox). .This .is .because .of .Newton's .third .law .that .states .applied .forces .acting .on

.one .another .will .be .equal .in .magnitude .and .opposite .in .direction. .




b) .The .change .in .momentum .of .both .the .car .and .the .mailbox .would .be .the .same
.according .to .the .conservation .of .momentum. .This .is .because .there .are .no .outside .forces

.acting .on .the .system .that .is .colliding .(the .car .and .the .mailbox). .This .is .why . we .use .the

.equation ."mv .= .mv". .




c) .The .change .in .velocity .of .the .mailbox .would .need .to .be .larger .than .the .change .in
.velocity .of .the .car. .This .is .because .for .the .systems .to .be .equal, .they .(obviously) .must

.equal .out. .So, .since .the .car .has .a .much .larger .mass .than .the .mailbox, .the .mailbox .must

.have .a .larger .change .in .velocity.




LESSON .9 .QUIZ .- .correct .answer. . . . .starting .now

What .maximum .horizontal .force .can .a .student .apply .to .the .top .of .a .2.0 .m .cubic .box .at .rest
.on .a .rough .surface .before .the .box .will .start .to .tip .over? .The .mass .of .the .box .is .100 .kg. .-

.correct .answer. . . . .490 .N




A .wood .board .with .a .length .of .2.5 .m .is .allowed .to .pivot .at .its .midpoint. .A .60 .kg .student .sits
.0.8 .m .from .one .end .of .the .board. .What .is .the .torque .of .the .student .on .the .board? .- .correct

.answer. . . . .265 .m-N




A .60 .kg .student .on .the .end .of .a .1.5 .m .diviner's .board. .What .is .the .student's .torque .on .the
.diving .board? .- .correct .answer. . . . .882 .m-N




Which .of .the .following .is .NOT .an .example .of .an .applied .net .torque? .- .correct .answer. . . . .A
.student .pushes .a .box .causing .it .to .slide .along .the .ground.




What .mass .m2 .will .allow .the .following .mobile .to .maintain .static .equilibrium? .- .correct
.answer. . . . .0.075 .kg

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