PHYSICS S1 REVIEW

ALL QUIZZES/TESTS MCQ

YOU TRAVEL 10 M/S FOR 2 SECONDS. THEN YOU SPEED UP AT A CONSTANT RATE FOR THE NEXT 4 SECONDS AND REACH 16 M/S

WHAT IS THE DISTANCE YOU TRAVELED?

YOU TRAVEL 10 M/S FOR 2 SECONDS. THEN YOU SPEED UP AT A CONSTANT RATE FOR THE NEXT 4 SECONDS AND REACH 16 M/S

WHAT IS THE DISTANCE YOU TRAVELED?

S=D/T →

10=D/2 → 10X2 = 20 →

D = 20M

S = D/T →

16=D/4 → 16X4 = 64

64 + 20 = 84 M TOTAL

YOURE PLAYING CATCH WITH YOUR DOG. THE DOG RUNS 15M AWAY FROM YOU IN 3 SECONDS. IT THEN COMES RIGHT BACK IN 3 SECONDS.

WHAT IS YOUR DOGS SPEED?

S = TOTAL D/TOTAL T →

S = 30/6 →

S = 5 M/S

YOURE PLAYING CATCH WITH YOUR DOG. THE DOG RUNS 15M AWAY FROM YOU IN 3 SECONDS. IT THEN COMES RIGHT BACK IN 3 SECONDS.

WHT IS YOUR DOG’S VELOCITY?

V = DISPLACEMENT/TIME →

15-15/6 = 0 →

VELOCITY = 0 M/S

YOU DROP A PENCIL OFF OF THE SECOND FLOOR OF RMHS. YOU MEASURE THE HEIGHT TO BE 16M.

HOW LONG WILL IT TAKE TO REACH THE GROUND?

D = VO(T) + ½ AT² →

D= ½ AT² → 2D = AT²→ √2D/A = T

√2(16)/-10 = T →

T = 1.79 SEC

A PENNY IS DROPPED INTO A WELL AND IT TAKES 4.2 SECONDS TO HIT THE GROUND.

HOW DEEP IS THE WELL?

D = VOT+1/2AT²

D = 0 + ½ (-10)(4.2)²

D= 88.2 M

A BALL IS THROWN STRAIGHT UP INTO THE AIR AT A VELOCITY OF 16 M/S

WHAT IS THE MAXIMUM HEIGHT? (dy)

VF² = VOY + 2ADY →

0 = (16)² + 2(-10)Y →

-256 = -20Y →

Y = 12.8

A BALL IS THROWN STRAIGHT UP INTO THE AIR AT A VELOCITY OF 16 M/S

HOW LONG WILL IT TAKE TO REACH THE MAXIMUM HEIGHT? (HALF OF TOTAL TIME)

VF = VO + AT

0 = 16 + -10T → -16 = -10T →

T=1.6 SECS

TOTAL TIME: 3.2 SECONDS

A ROCK IS THROWN OFF A CLIFF THAT IS 24 M HIGH W/ A HORIZONTAL VELOCITY OF 8 M/S

HOW LONG WILL IT TAKE TO HIT THE GROUND?

D= VOY + ½ AT² → √2D/A = T

√2(24)/-10 = 2.19 SECONDS

A ROCK IS THROWN OFF A CLIFF THAT IS 24 M HIGH W/ A HORIZONTAL VELOCITY OF 8 M/S

WHAT IS THE RANGE OF THE BALL (HOW FAR WILL IT LAND)

DX = VX(T)

D=8X2.19 → DX = 17.53M

YOU KICK A SOCCER BALL AT AN ANGLE OF 45 DEGREES AT A VELOCITY OF 20 M/S

SOLVE FOR VY AND VX

VY: SIN(45) X 20 = 14.14

VFY = VO + AT

= (14.14) + -10T → -14.14 = -10T → T = 1.41 SECONDS

VX: COS(45) X 20 = 14.14

YOU KICK A SOCCER BALL AT AN ANGLE OF 45 DEGREES AT A VELOCITY OF 20 M/S

HOW HIGH WILL THE GOLF BALL GET?

DY = VOT + ½ AT²

= 14.14(1.41) + ½ (-10)(1.41)² →

DY = 10 M

YOU KICK A SOCCER BALL AT AN ANGLE OF 45 DEGREES AT A VELOCITY OF 20 M/S

HOW LONG WILL THE BALL BE IN THE AIR (TOTAL TIME)

1.41 X 2 =

2.83 SECONDS

YOU KICK A SOCCER BALL AT AN ANGLE OF 45 DEGREES AT A VELOCITY OF 20 M/S

HOW FAR WILL THE BALL GO?

DX = VX(T)

= 14.14 X 2.83

DX = 40.02 M

BONUS QUESTION: A FOOTBALL PLAYER KICKS A GOOTBALL AT AN INITAL SPEED OF 25 M/S AT AN ANGLE OF 40 DEGREES ABOVE THE HORIZONTAL. THE GOALPOST IS 3.05 M HIGH AND IS 25 M AWAY FROM THE KICKER.

HOW LONG IS THE FOOTBALL IN THE AIR BEFORE IT REACHS THE CROSSBAR?

[STILL HAVE TO SOLVE CORRECTLY]

BONUS QUESTION: A FOOTBALL PLAYER KICKS A GOOTBALL AT AN INITAL SPEED OF 25 M/S AT AN ANGLE OF 40 DEGREES ABOVE THE HORIZONTAL. THE GOALPOST IS 3.05 M HIGH AND IS 25 M AWAY FROM THE KICKER.

WHAT IS THE FOOTBALLS HEIGHT WHEN IT REACHES THE CROSSBAR (25 METERS AWAY HORIZONTALLY)

[STILL HAVE TO ANSWER CORRECTLY]

BONUS QUESTION: A FOOTBALL PLAYER KICKS A GOOTBALL AT AN INITAL SPEED OF 25 M/S AT AN ANGLE OF 40 DEGREES ABOVE THE HORIZONTAL. THE GOALPOST IS 3.05 M HIGH AND IS 25 M AWAY FROM THE KICKER.

WILL THE FOOTBALL CLEAR THE CROSSBAR?

YES

WHY: [IDK; INTIALLY SAID NO BUT IT WAS WRONG]

IF THE NET FORCE ON THE BOX ABOVE IS 160 N, HOW MUCH FRICTION IS THERE

90 N

IF THE APPLIED FORCE IN THE DIAGRAM WAS DIRECTED AT AN UPWARD ANGLE, WHAT WOULD HAPPEN TO THE MAGNITUDE (AMOUNT) OF THE NORMAL FORCE?

IT WOULD DECREASE

BRIAN HAS A MASS OF 75KG ON EARTH WHERE GRAVITY IS -10 M/S². WHAT WOULD HIS MASS BE N THE MOON WHERE GRAVITY IS ONLY 1.5 M/S²

75 KG

WHY: MASS NEVER CHANGES

A BUG FLIES IINTO THE PATH OF A TRUCK AND IS STRUCK BY THE WINDSHIELD. THE IMPACT FORCE IS

SAME FOR BOTH

WHY: NEWTONS 3RD LAW — ORIGINAL ANSWER: GREATER ON THE BUG: ONLY TRUE WHEN ASKING FOR ACCELERATION

WHICH TWO OF THE FOLLOWING STATEMENTS ARE TRUE?

THE SUM OF THE FORCES ACTING ON AN OBJECT AT REST MUST BE ZERO

AN OBJECT MOVING AT A CONSTANT VELOCITY MUST HAVE A NET FORCE OF ZERO

WHAT IS THE MASS OF THE SKYDIVER?

83.3 KG

IN DIAGRAM B ABOVE, WHAT IS THE NET FORCE ON THE SKYDIVER

483 N

IN WHICH DIAGRAM DOES THE SKYDIVER HAVE THE GREATEST ACCELERATION?

DIAGRAM A

DETERMINE THE ACCELERATION OF THE BOXES IF THEY ARE PULLED WITH 25 N OF FORCE

1.3 M/S²

AN AIRLINE PILOT REALIZES THAT SHES LATE FOR HER NEXT FLIGHT. SHE BEGINS RUNNING DOWN THE TERMINAL, ACCELERATING AT 2 M/S² AS SHOWN IN THE PICTURE BELOW. WHICH OF THE FOLLWING FORCE VECTORS SHOWS THE NET FORCE ACTING ON THE SUITCASE

D - MOVING RIGHT

A BOX WITH A WEIGHT OF 50 N IS AT REST ON A HORIZONTAL SURFACE. THE COEFFICIENT OF STATIC FRICTION BETWEEN THE BOX AND THE SURFACE IS 0.50, AND THE COEFFICIENT OF KINETIC FRICTION IS .30. A HORIZOTAL FORCE OF 20 N IS THEN EXERTED ON THE BOX. THE MAGNITUDE OF THE ACCELERATION OF THE BOX IS MOST NEARLY

0 M/S²

A 10KG BLOCK IS HELF AT REST BY TWO STRINGS, WITH TENSIONS T1 AND T2. WHICH EXPRESSION IS CORRECT ABOUT THE TENSION IN THE STRINGS

T1<T2

T2 IS A SHORTER STRING THEREFORE IT REQUIRES MORE TENSION

IDENTIFY THE FOLLOWING IMAGE

GLOVE PUSHES BALL RIGHTWARDS

A 45KG PERSON STANDS ON A SCALE IN AN ELEVATOR. THE SCALE READS 460N. WHAT DOES THIS TELL YOU ABOUT THE MOTION OF THE ELEVATOR

IT IS MOVING UPWARDS, ACCELERATING UPWARDS

WHY; 460N IS BIGGER THAN 450 (45X10=450)

IN PREPARTION FOR A BIG DELIVERY, PETE A PIZZA DELIVERY GUY PULLS A 20 KH CASE OF PIZZAS ACROSS A ROOM WITH A FORCE OF 250N AT AN ANGLE OF 20 DEGREES ABOVE THE HORIZONTAL HE P LACES THEM ON HIS SKATEBOARD SO WE CAN ASSUME THERE IS NO FRICTION

FIND THE HORIZONTAL (FX) AND VERTICAL (FY)

FY: SIN(20) X 250 = 85.51

FX = COS(20) X 250 = 234.92

IN PREPARTION FOR A BIG DELIVERY, PETE A PIZZA DELIVERY GUY PULLS A 20 KH CASE OF PIZZAS ACROSS A ROOM WITH A FORCE OF 250N AT AN ANGLE OF 20 DEGREES ABOVE THE HORIZONTAL HE P LACES THEM ON HIS SKATEBOARD SO WE CAN ASSUME THERE IS NO FRICTION

FIND THE ACCELERATION OF THE CASE OF PIZZAS

FN = MA

234.92/20 = A

A = 11.75 M/S²

IN PREPARTION FOR A BIG DELIVERY, PETE A PIZZA DELIVERY GUY PULLS A 20 KH CASE OF PIZZAS ACROSS A ROOM WITH A FORCE OF 250N AT AN ANGLE OF 20 DEGREES ABOVE THE HORIZONTAL HE P LACES THEM ON HIS SKATEBOARD SO WE CAN ASSUME THERE IS NO FRICTION

HOW COULD PETE INCREASE THE ACCELERATION OF THE CASE OF PIZZAS, ASSSUMING 250N IS PETES MAX PULLING FORCE

HE COULD CARRY THE PIZZAS AT A LOWER ANGLE LIKE 15 DEGREES

PETE THE PIZZA DELIVERY GUY WITH A 20 KG PIZZA IN HANFD,NOW RIDES AN ELEVATOR UP TO THE 12TH FLOOR OF THE BUILDING. AS THE ELEVATOR STARTS TO ASCNED IT ACCELERATES 3 M/S² UPWARD

CALCULATE THE NORMAL FORCE ACTING ON HIM WHILE HE ACCELERATES

60 + 20 = 80

80 X 3 =

240 N

PETE NOW NEEDS TO PASS THE PIZZA ACROSS A LONG TABLE TO A GROUP OF HUNGRY TEENAGERS. CONSDIER FRICTIONAL FORCE ON TABLE

IF THE PIZZA HAS A MASS OF 20KG AND PETE PUSHES WITH AN APPLIED FORCE OF 80N DETERMINE THE ACCELERATION OF THE PIZZA. THE COEFFICIENT OF FRICION IS .3

A = 1 M/S²

[WORK DOESNT MAKE SENSE ON PAPER]

WHAT TYPE OF COLLISION WERE CART 1 AND CART 2 INVOLVED IN

ELASTIC

AT WHAT TIME DID THE CARDS COLLIDE

5 SECONDS

THE RATIO OF THE MASSES OF CART 1 TO CART 2 IS

1:2

WHY: CART 1 CHANGE IN VELOCITY: 11 M/S

CART 2 CHANGE IN VELOCITY: 5.5 M/S

AFTER THE COLLISON CART 1 VELOCITY

INCREASED

THE TOTAL MOMENTUM OF THE TWO CART SYSTEM

REMAINED THE SAME AFTER THE COLLISON

WHY: MOMENTUM IS CONSERVED IN COLLISIONS

IN WHICH EXPERIMETN DID THE ART HAVE A GREATER IMPULSE

BOTH HAD THE SAME

IN EXPERIEMENT 2 THE CART EXPERIENCES A SMALLER FORCE IN STOPPING IT, THIS IS DUE TO

LONGER IMPACT TIME

THE CHANG EIN MOMENTUM OF THE CART IN EXPERIEMTN 1 COMPARED TO THE CHANGES IN MOMENTUM IS CART 2 IS

THE SAME

MOMENTUM IS CONSEREVED

THE RECOIL VELOCITY OF A RIFLE IS LESS THAN THE VELOITY OF THE BULLET BECAUSE

RIFLE HAS MUCH MORE MASS THAN BULLET

IF A MONKEY FLOATING OUTER SPACE TRHOWS HIS HAT AWAY, THE HAT AND THE MONKEY

MOVE AWAY FROM EACHOTHER AT DIFFERNT SPEEDS

IMPULSE IS ALWYAS EQUAL TO THE CHANGE IN MOMENTUM

TRUE

A MOUSE AND ELEPHANT CAN NEVER HAVE THE SAME MOMENTUM

FALSEA

WITHOT FRICTION, MOMENTUM IS CONSEREVED IN EVERY COLLISION

TRUE

OBJECT INVOLVED IN A COLLISION ALWAYS EXPERINES AN ACCELERATION

TRUE

IN AN INELASTIC COLLISION THE OBJECT WILL ALWAYS SLOW DOWN

FALSE

A 4K BALL HAS A MOMENTUMOF 12 KHM/S. WHATS THE SPEED?

3

12=4V

V=3

A BALL IS MOVING AT 3 M/S AND HAS A MOMENTUM OF 48 KGM/S. WHATS THE BALLS MASS?

48=3M

M=16

A .25KG BLOB OF CLAY MOVING AT 4 M/S STICKS TO A WALL EXPERICING A FORCE OF 10 N. HOW LING DID IT TAKE THE BLOB TO STOP?

FXT=MX CHANGE IN V

10T = .25(-4)

T = .1 SEC

A 1000 KG CAR MOVING AT 10 M/S BRAKES TO STOP IN 5 SECONDS. THEA VERGAE BREAKING FROCE IS

2000 N

FXT = M X CHAIGE IN V

5F = 1000 X -10

5F = -10,000

F = 2,000N

A 5KG FISH DWIMMING IN 1 M/S SWALLOS AN ABSENT MINDED 1 KG FISH AT REST THE SPEED OF THE LARGER FISH AFTER HIS LUNCH IS

M1V1+M2V2 = M1V1 + M2V2

5(1) + 1(0) = 5V + 1

5 = 6V

V = 5/6 M/S

A 10,000 KG RAILROAD CAR TRAVELING AT 24 M/S STRIKES AN IDENTICAL CAR INITALLY AT REST IF THE CARS LOCK TOGETHER AS A RESULT OF THE COLLISION, WHATS THIER COMBINED SPEED?

M1V1+M2V2 = (M1+M2)V

240,000 = 2000V

V = 12 M/S

A BOWLING BALL W A MASS OF 5 KG GOING A SPEED OF 8 M/S STRIKES A PIN W A MASS OF 1.2 KG AT REST. THE PIN ENDS UP GOING 10 M/S. WHATS THE FINAL SPEED OF THE BALL AFTER THE COLLISION?

M1V1+M2V2= M1V1+M2V2

40 = 5V + 12

28 = 5V

V = 5.6 M/S

TWO SKATERS ARE STANDING IN THE MIDDLE OF THE ICE RINK AT THE BEGINNING OF A ROUTINE PABLO HAS A MASS OF 65KG AND ISA HAS 50KG. AFTER THEY PUSH OFF, ISA IS MOVING BACKWARDS AT 3 M/S. HOW FAST IS PABLO GOING?

(M1+M2)V = M1V1 + M2V2

V = 2.31 M/S

A 12KG HAMMER STRIKES AT AL NAIL AT A VELOCITY OF 8.5 M/S AND COMES TO REST IN A TIME INTERVAL OF 0.008 SECONDS.

WHAT IS THE IMPULSE GIVEN TO THE NAIL

M X CHANGE IN V

12 X 8.5 =

102 J

A 12KG HAMMER STRIKES AT AL NAIL AT A VELOCITY OF 8.5 M/S AND COMES TO REST IN A TIME INTERVAL OF 0.008 SECONDS.

WHAT IS THE AVERGAE FORCE ACTING ON THE NAIL

F X T = 102

F X .008 = 1-2

F = 12,750 N

EGG DROP

GROUP A : PLASTIC CUP W ONE SINGLE PAPER AS PARACHUTE

GROUP B: STRAWS

STRAWS BECUASE ITLL SLOW DOWN THE TIME OF IMPACT AND HAVAE A LESSER FORCE IMPACT ON EGG

AT WHICH OF THE FOLLOWING WAS THE ROLLER OCOASTER TRAVELLING THE FASTEST?

22 SEC

WAHT WSA THE APPROXIMATE VELOCITY OF THE ROLLER COASTER AT 10 M/S

13 M/S

87500 = 500V²

√175 = V²

ABOUT HOW HIGH WAS THE ROLLER COASTER AT THE BEGINNING OF THE RIDE

40 M

PE = MGH

1100 X 10 X H

440000 = 11000H

H = 40 M

WHAT WAS THE TOTAL MECHANICAL ENERGY OF THE ROLLER COASTER DURING THIS RIDE

440,000J

WHAT IS SIGNIFICANT ABOUT THE PROTION OF THE GRAPH WHERE THE 2 PLOTS OVERLAP AT 13.5-15 SECONDS

KINETIC ENERGY AND POTENTIAL IS THE SAME

POWER IS DEFINED AS

WORK DONE ON AN OBJECT DIVIDED BY THE TIME TAKEN TO DO THE WOKR

YOU WALK UP THE STIARS ONE DAY AND TEHN RUN UP THE NEXT DAY YOUR AMOUNT OF __ IS DIFFERENT

POWER

JOULES IS THE UNIT FOR

WORK, KE, PE

IF MR BATEY PUSHES AN OBJECT WITH TICE THE FORCE FOR TWICE THE DISTANCE HE DOES

FOUR TIMES THE WORK

AN APPLE FALLS OFF THE DESK. BEFORE IT FELL IT HAD 20J. F PE PART WAY DOWN IT HAS 5 J. PE. THIS MEANS KE IS

15 J

20-5 =15

A PENDULUM SWINGS BACK AND FORTH WHAT POINT AHS THE GREATEST SOEED

POINT B (THE MIDDLE)

WHAT CAN YOU SYA ABOUT POINT C (RIGHT SIDE)

NO KE

A ROLLER COASTER GENEREALLY GENERATES BY

PE AND KE BEIGN TRANSFERRED

IN AN IDEAL SITUATION, A BOUNCY BALL IS DROPPED HITS THE FLOOR AND BOUCNES BACK UO. WHICH OF THE FOLLWOING SHOULD BE TRUE

IT QILL RETURN TO ITS ORIGNAL HEIGHT

IF YOU TRIPLE THE VELOCITY OF A MOVING OBJECT, WHAT HAPPENS TO KE

9X THE ORIGINAL VALUE

PE IS THE ENRGY AN OBJECT HAS BECAUSE OF ITS

LOCATION

(HEIGHT)

WHICH REQUIRES MORE WORK: 70KG SACK 2 METERES VERTICALLY OR 35 SACK 4 METERS VETICALLY

SAME AMOUNT OF WORK

BOTH EQUAL 140

STIEVE RUNS UP A FLIGHT OF STAIRS 8 METERS TALL IN 4 SECONDS. IF STEIVE HAS A FORCE OF 980 NETWONS APPROXIMATELY HOW POWERFUL IS HE?

2000 W

980X8 / 4 = 1960

A BALL IS TRHOWN STRIAGHT UP INTO THE AIR. WHICH STATEMENT DESCRIBES ENERGY TRANSFORMATION AS THE BALL RISES

KE DECREASES PE INCREASES

AN ARROW W A MASS OF .5 KG IN A BOW OF 70J HAS PE ASSUM ING NO HEAT LOSS, HOW MUCH KE REMAINED?

70J

WHY: NO HEAT LOSS

YOU THROW A ,4 KG BALL STRAIGHT UP INTO THEA IR. WHEN IT LEAVES YOUR AHDN ITS MOVING AT 3 M/S.

WHAT IS THE PE AT TOP OF ITS PATH

KE = ½ (.4)(3)²

= 1.8J

HOW MUCH WORK DOES IT TAKE TO THROW THE BALL UP INTO THE AIR

PE = MGH

.4 = .2(10)(H)

= .2

A ROLLER COASTER TRAVELS DOWN A LARGE HILL. IT GAINS 850J OF KINETIC ENERGY AS IT TRAVELS 5 M/S. WHT IS THE MASS OF THE ROLLER COASTER

KE = ½ MV²

850 = ½ M (5)²

=68 KG

HOW MANY JOULES OF ENERGY ARE USED BY A 450 W MOTOR THAT RUNS FOR 1.5 HOURS?

1.5X60X60 = 5400 SECONDS (CONVERT 1.5 HOURS TO SECONDS)

450 = W/5400

W = 2,430,000 J