PHY-2048C – Fall 2020 SI with Camilo

Study Union Review

Disclaimer: This practice test does not necessarily cover all the material that is going to appear on the test, the questions are not the same as the actual test, and should not be used as the sole study guide for the test.

Topics:

2D Motion Forces Circular motion Energy and Work Linear momentum and conservation of momentum Rotational kinematics Angular momentum Simple harmonic motion

1. A batter hits a baseball so that it leaves the bat at speed V 0 = 37m/s at and angle of 53.1 degrees.

a) Find the velocity (magnitude and direction) at t = 2 seconds

b) Find the time when the ball reaches its highest point and the height at this time

c) Find the horizontal distance that the ball travels as it hits the ground and the velocity right before impacting the ground.

V x=V 0 cos (53.1 )=37cos (53.1 )=22.2m / s

V y0=V 0sin (53.1 )=37 sin (53.1 )=29.6m / s

V f=V 0+a∗t

V y f=V y0+a∗t=29.6−2∗9.8=10m / s

V f=√V y f

2+V x f

2=√102+22.22=24.4m / sθ=arctan (10 / 22.2 )=24.2°

b) Find the time when the ball reaches its highest point and the height at this time

V f=V 0+a∗t

0=29.6+(−9.8)∗t

t=29.6 / 9.8=3.02 s

y=V 0t+12a∗t2

y=29.6∗3.02−129.8∗3.022=44.7m

c) Find the horizontal distance that the ball travels as it hits the ground and the velocity right before impacting the ground.

x=V 0 t+120∗t 2

x=V x t=22.2∗6.04=134.088m

V x=22.2m / s

V y0=−29.6m / s

V f=√V y f

2+V x f2=37m /s

2. A box (m1) is on a 30-degree inclined ramp. This mass is connected by a string to another mass (m2) that is suspended over a massless and frictionless pulley. Mass 1 is 10 kg; Mass 2 is 30 kg. The coefficient of kinetic friction of Mass 1 and the inclined surface is 0.2.

a) Find the acceleration of the system. b) Find the tension in the rope.

3. A force of 130 N is applied to the front of a sled at an angle of 27.0 above the horizontal so as to pull the sled a distance of 170 meters. How much work was done by the applied force?

Ans. W = F •d cos = 130 N • 170 m cos 27.0 = 19691.244 J

4. A block of mass m1 = 20.0 kg is connected to a block of mass m2 = 35.0 kg by a massless string that passes over a light, frictionless pulley. The 35.0-kg block is connected to a spring that has negligible mass and a force constant of k = 222 N/m as shown in the figure below. The spring is unstretched when the system is as shown in the figure, and the incline is frictionless. The 20.0-kg block is pulled a distance h = 25.0 cm down the incline of angle θ = 40.0 ◦ and released from rest. Find the speed of each block when the spring is again unstretched.

v=√2∗9.8∗0.25∗(35−20sin(40 °))+222∗¿¿¿

5. A 10.0kg block is released from rest at point A in the figure below. The track is frictionless except for the portion between points B and C, which has a length of 6.00 m. The block travels

down the track, hits a spring of force constant 2 200 N/m, and compresses the spring 0.200 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between points B and C.

We choose the horizontal surface as the 0-potential energy. Between A and B, energy is conserved and it amounts to : mgh = (1/2)mVB

2

Between B and C, energy is not conserved due to friction. The change in energy is the work done by friction : (1/2)mVc

2 − (1/2)MvB2 = fk · d = −µN d = −µkmgd

Remember N = mg from 2nd Law.

Between C and the final position, energy is conserved and it amounts to (1/2)mVc2 = (1/2)

k(∆x)2 ; ∆x being the spring compression distance.

6. A satellite orbits Earth at an altitude of 400 kilometers above the planet’s surface.

(radius of earth = 6.38x106m)

a) What is its speed in meters per second?

b) What is the period of the satellite?

a)

h=400 km=400 km∗1000m1km

=4 x105m

r=rearth+h=6.38 x106m+4 x105m=6.78 x106m

v=√GM earth

r=√GM earth

r

b)

v=√GM earth

r=2πr

T

T=2π √ r3

GM earth=2 π √ (6.78x 106)3

(6.67 x 10−11)∗(5.98 x1024)=5554.064 s

7. Two crates are sliding on a frictionless surface as shown in the figure below. The 11 kg crate is sliding to the right at 9.0 m/s and the 26 kg crate is sliding to the left at 6.0 m/s. The two crates collide and stick together. Use conservation of momentum to find the velocity of the two crates after the collision.

p0=p f

v=11∗9+26∗(−6)11+26

=−1.541m /s

8. A 0.530-kg basketball hits a wall head-on with a forward speed of 18.0 m/s. It rebounds with a speed of 13.5 m/s. The contact time is 0.100 seconds. (a) determine the impulse with the wall, (b) determine the force of the wall on the ball.

a)

impulse=J=∆ p=m∆v=mv f−mv0=(0.530kg)•(−13.5m /s−18.0m /s)

J=−16.7Ns

b)

J=Ft

F=J / t=(−16.7)/0.1=−167 N

9. Two masses are attached to a massless string that goes around a massive pulley as shown in the picture. Find an equation that gives the value of the acceleration of the system (a) in terms of m1, m2, M, and g.

10. Starting from rest, a 15-cm-diameter compact disk takes 4.0 s to reach an angular velocity of 3000 rpm. Assume that the angular acceleration is constant. The disk's moment of inertia 3x10-5kgm2.

a. How much torque is applied to the disk?

b. How many revolutions does it make before reaching full speed?

ωf=3000 rpm=3000 rev /m

2 πrad1rev

∗1m

60 s=314.16 rad / s

R = 0.075

α=ωf

t=314.16 rad / s

4 s=78.54 rad / s2

∑ τ=αI=(78.54 )∗3 x10−5=0.00236Nm

θ=ω0 t+12α∗t2

θ=12(78.54)∗42=628.319 rad

628.319 rad=628.319 rad 1 rev2 πrad

=100 rev

11. A classic physics demonstration involves firing a bullet into a block of wood suspended by strings from the ceiling. The height to which the wood rises below its lowest position is mathematically related to the pre-collision speed of the bullet. If a 9.7-gram bullet is fired into the

center of a 1.1-kg block of wood and it rises upward a distance of 33 cm, then what was the pre-collision speed of the bullet?

∆ E=∆ KE+∆PE=0

KE0=PE f

12(mwood+mbullet)v f

2=(mwood+mbullet)gh

12v f2=gh

v f=√2 gh=√2∗9.8∗0.33=2.5432m /s

mwood∗0+mbullet vbullet0=(mwood+mbullet)v f

(0.0097 kg)•(vbullet 0) = (1.1 kg + 0.0097 kg)•(2.5432 m/s)

(0.0097 kg)•( vbullet 0) = 2.8222 kg•m/s

vbullet 0 = (2.8222 kg•m/s) / (0.0097 kg)

vbullet 0 = 290.95 m/s

vbullet 0 = ~2.9 x 102 m/s

12. A typical small rescue helicopter has four blades: Each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. 

(a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm.

∑ I=4∗( 13mblade(r blade)

2)=4∗( 13∗50∗(4)2)=1066.667 kg (m)2

ω=

300 revmin

∗1min

60 sec∗2 πrad

1 rev=10πrad / s

KEr=12I ω2=0.5∗1066.667∗(10π2)=526378.9 J

13. A 25-kg child stands at a distance r=1.0m from the axis of a rotating merry-go-round. The merry-go-round can be approximated as a uniform solid disk with a mass of 500 kg and a radius of 2.0 m. Find the moment of inertia of this system.(treat the child as a point mass)

I total=I child+ Imerry=mchild rchild2+ 12mmerry rmerry

2

I total=25∗12+0.5∗500∗22=1025 kgm2

14. A uniform ladder is L=5.0m long and weighs 400.0 N. The ladder rests against a slippery vertical wall, as shown in the figure. The inclination angle between the ladder and the rough floor is β=53°. Find the reaction force from the wall on the ladder and the coefficient of static friction μs at the interface of the ladder with the floor that prevents the ladder from slipping.

Fnety=N floor−W=0

N floor=W=400 N

Fnetx=F f−Nwall=0

Nwall=F f=μsN floor

I decide to let the axis of rotation to be on the lower end of the ladder to make the torques of the forces from floor equal 0

τ net=τweight−τnormal force of thewall=400(2.5)sin(37)−N wall(5)sin(53)=0

Nwall=150.7N

Nwall=μsN floor

μs=Nwall

N floor=0.377

15. A rod of length L = 1.0 m and mass M = 2 kg is hinged on one side so that it can rotate freely(imagine like a door but it is a rod). A bullet is then fired and hits the rod right at the edge. The bullet has mass m=5g and a speed of 200 m/s. What is the angular velocity of the rod with respect to the hinge just after the bullets embeds itself in the door?

m = mass bullet M = mass of rod

L0=LF

L=Iω

L0=Lbullet 0+Ldoor 0

Ldoor0=0

L0=I bullet❑ωbullet❑=m R2ω=mR(Rω)=mRv

Rω=v

LF=Lbullet∧door f=I ω=(mR2+ 13M R2)ωf

mRv=(mR2+ 13M R2)ωf

mRv

(m R2+ 13M R2)

=ωf=(0.005)∗200

(0.005+ 132)

=1.5rad /s

16. Consider a 550 g ball hanging on a spring with a spring constant equal to 22 N/m. We pull down on the ball and let it go, so that it oscillates up and down. Find:

(a) The period of the oscillation.

(b) The frequency of the oscillation.

T=2π √mk =2 π √ 0.5522 =0.993 s

f= 1T

= 10.993

=1.00638Hz

17. Two balls each of unknown mass m are mounted on opposite ends of a 1.5-m-long rod of mass 550 g. The system is suspended from a wire attached to the center of the rod and set into torsional oscillations.

If the torsional constant of the wire is 0.63 N⋅m/rad and the period of the oscillations is 5.6 s, what is the unknown mass m?

k = 0.63Nm/rad     T=5.6s         L=1.5m      M=0.55kg