PHY-2048C – Spring 2020 SI with Camilo

Exam 2 Practice Test

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

1. A force of 120 N is applied to the front of a sled at an angle of 28.0 above the horizontal so as to pull the sled a distance of 165 meters. How much work was done by the applied force?

Ans. W = F •d cos = 120 N • 165 m cos 28.0 = 17,482.36 J

2. A car is sitting at the top of an inclined plane, which is 5.2 meters long and meets the horizontal at an angle of 12.0º. The cart is then allowed to roll to the bottom of the incline. What will be the velocity of the cart as it reaches the bottom of the incline?

v = √𝟐𝒈𝒉 = √𝟐 (𝟗. 𝟖 𝐦 𝐬 𝟐 )(𝟏. 𝟎𝟖 𝒎) = 4.6 m/s

3. Two objects are connected by a light string passing over a light, frictionless pulley as shown in the figure below. The object of mass m1 = 6.60 kg is released from rest at a height h = 3.40 m above the table. Using the isolated system model, determine the speed of the object of mass m2 = 3.00 kg just as the 6.60kg object hits the table.

4. 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.

5. A block of mass m1 = 18.0 kg is connected to a block of mass m2 = 32.0 kg by a massless string that passes over a light, frictionless pulley. The 32.0-kg block is connected to a spring that has negligible mass and a force constant of k = 220 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 18.0-kg block is pulled a distance h = 24.0 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.

6. A golf player hits a 0.045kg golfball that is initially at rest, changing its momentum by 4.2 (kg m)/s What is the final speed of the golfball?

pchange=p f−p0

pchange=mV f−0

V f=pchangem

= 4.20.045

=93.333m /s

7. Two crates are sliding on a frictionless surface as shown in the figure below. The 10 kg crate is sliding to the right at 8.0 m/s and the 25 kg crate is sliding to the left at 5.0 m/s. The two crates collide and stick together. Use conservation of momentum to find the velocity of the two crates after the collision.

8. A 10 kg mass traveling 2 m/s meets and collides elastically with a 2 kg mass traveling 4 m/s in the opposite direction. Find the final velocities of both objects.

(I am adding a really detailed explanation on how to solve the equations here, so, if you want the answers you can just scroll down.)

First, we know total momentum is conserved in the collision.total momentum before collision = total momentum after collisionmAVAi + mBVBi = mAVAf + mBVBf

mAVAi – mAVAf = mBVBf – mBVBi

mA(VAi – VAf) = mB(VBf – VBi)Let’s call this Equation 1 and come back to it in a minute.

Since we were told the collision was elastic, the total kinetic energy is conserved.kinetic energy before collision = kinetic energy after collection½mAVAi

2 + ½mBVBi2 = ½mAVAf

2 + ½mBVBf2

mAVAi2 + mBVBi

2 = mAVAf2 + mBVBf

2

mAVAi2 – mAVAf

2 = mBVBf2 – mBVBi

2

mA(VAi2 – VAf

2) = mB(VBf2 – VBi

2)Use the “difference between two squares” relationship (a2 – b2) = (a + b)(a – b) to factor out the squared velocities on each side.mA(VAi + VAf)(VAi – VAf) = mB(VBf + VBi)(VBf – VBi)

Now we have two equations and two unknowns, VAf and VBf.Divide this equation by equation 1 from before (the total momentum equation from above) to get

Now we can cancel out most of this

This leavesVAi + VAf = VBf + VBi

Solve for VAf

VAf = VBf + VBi – VAi

Now we have one of our unknowns in terms of the other unknown variable. Plug this into the original total momentum equationmAVAi + mBVBi = mAVAf + mBVBf

mAVAi + mBVBi = mA(VBf + VBi – VAi) + mBVBf

Now, solve this for the final unknown variable, VBf

mAVAi + mBVBi = mAVBf + mAVBi – mAVAi + mBVBf

subtract mAVBi from both sides and add mAVAi to both sidesmAVAi + mBVBi – mAVBi + mAVAi = mAVBf + mBVBf

2mAVAi + mBVBi – mAVBi = mAVBf + mBVBf

factor out the masses2 mAVAi + (mB – mA)VBi = (mA + mB)VBf

Divide both sides by (mA + mB)

Now we know the value of one of the unknowns, VBf. Use this to find the other unknown variable, VAf. Earlier, we foundVAf = VBf + VBi – VAi

Plug in our VBf equation and solve for VAf

Group the terms with the same velocities

The common denominator for both sides is (mA + mB)

Be careful of your signs in the first half of the expressions in this step

Now we’ve solved for both unknowns VAf and VBf in terms of known values.

Now we put our values:

VAf = 0 m/sThe final velocity of the larger mass is zero. The collision completely stopped this mass.Now for VBf

Plug in our known values

VBf = 6 m/s

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.