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Question 1

Impulse and Momentum

Impulsive Force— a large force acting for a short period of time

The impulse of a force (I) is defined as the product of the force and the time during which it acts.

$$ I = F \times t $$

Where:
  I → impulse (N·s)
  F → force (N)
  t → time in seconds

Only external forces give impulses to a body. The direction of the impulse is the same as the direction of the force involved.
Momentum is a quantity of motion. It is the property of a body that determines the time required to bring it to rest. It is the product of its mass m and instantaneous velocity v.

$$ \text{Momentum} = m \times v $$

5. The unit of momentum derived from mass and velocity is kg·m/s.

Answer

The fastest cooling direction is opposite the gradient. For $T=xyz(1-x)(2-y)(3-z)$, at $(1/2,1,1)$ the $x$ and $y$ partials are zero by symmetry. The $z$ partial is:
$T_z=xy(1-x)(2-y)(3-2z)$
At $(1/2,1,1)$:
$T_z=(1/2)(1)(1/2)(1)(1)=1/4$
Opposite gradient is:
$\boxed{-\frac{1}{4}k}$

Answer

The coefficient of restitution relates rebound to drop height through velocity. Since $v = \sqrt{2gh}$ and $e = v_2/v_1$:
$$h_2 = e^2 h_1 = (0.60)^2(2.6)$$
$$\boxed{h_2 = 0.936 \text{ m}}$$

Answer

Part 1.

Using $v^2 = 2gh$ for the free fall onto the floor:
$$h = \frac{v^2}{2g} = \frac{12^2}{2(9.81)} = \boxed{7.34 \text{ m}}$$

Part 2.

Rebound velocity from the coefficient of restitution:
$$v' = e\,v = 0.85(12) = \boxed{10.2 \text{ m/s}}$$

Part 3.

Rebound height from the rebound velocity:
$$h' = \frac{v'^2}{2g} = \frac{10.2^2}{2(9.81)} = \boxed{5.3 \text{ m}}$$

Answer

Part 1.

Conservation of momentum (perfectly plastic): $v = \dfrac{m_Av_A + m_Bv_B}{m_A + m_B}$. For the x-component:
$$v_x = \frac{10(12) + 15(6)}{25} = \frac{210}{25} = \boxed{8.4 \text{ m/s}}$$

Part 2.

For the y-component:
$$v_y = \frac{10(-23) + 15(12)}{25} = \frac{-50}{25} = \boxed{-2 \text{ m/s}}$$

Part 3.

For the z-component:
$$v_z = \frac{10(14) + 15(16)}{25} = \frac{380}{25} = \boxed{15.2 \text{ m/s}}$$

Answer

For rolling resistance of a wheel pulled at the center,
$$P=W\frac{e}{r}$$
Thus,
$$e=\frac{Pr}{W}$$
With $r=300/2=150\text{ mm}$,
$$e=\frac{(90)(150)}{18000}=0.75\text{ mm}$$
Therefore, the coefficient of rolling resistance is $\boxed{0.75\text{ mm}}$.

Answer

For rollers, the pull may be computed from the rolling-resistance offsets at the floor and at the load contact:
$$P=W\frac{e_1+e_2}{D}$$
Here, $W=1000\text{ kN}=1.0\times10^6\text{ N}$, $e_1=0.08\text{ mm}$, $e_2=0.02\text{ mm}$, and $D=50\text{ mm}$.
$$P=(1.0\times10^6)\frac{0.08+0.02}{50}=2000\text{ N}$$
Therefore, the required pull is $\boxed{2000\text{ N}}$.

Answer

Take right as positive. First find the second body's final velocity from momentum:
$100(5)+140(-3)=100(-2)+140v_2$
$80=-200+140v_2 \Rightarrow v_2=2\text{ m/s}$
Coefficient of restitution:
$e=\frac{v_2-v_1}{u_1-u_2}=\frac{2-(-2)}{5-(-3)}$
$\boxed{e=0.50}$

Answer

For a vertical rebound, the rebound height is related to the drop height by
$h_2=e^2h_1$
$h_2=(0.92)^2(6)$
$h_2=5.0784$
$\boxed{5.08}$

Answer

For impact with a smooth horizontal surface, the horizontal component is unchanged and the vertical component after impact is multiplied by $e$.
$\tan\theta=e\tan30^\circ$
$\tan\theta=0.80\tan30^\circ$
$\theta=\tan^{-1}(0.80\tan30^\circ)$
$\boxed{\theta=24.79^\circ}$

Answer

The ball first reaches the ground before it reaches the wall.
Time to first hit the ground:
$t_1=\sqrt{\frac{2(3)}{9.81}}=0.782\text{ s}$
Horizontal distance traveled before bounce:
$x_1=30(0.782)=23.46\text{ m}$
Vertical speed just before bounce is $\sqrt{2gh}=7.67\text{ m/s}$, so after bounce:
$v_y=0.70(7.67)=5.37\text{ m/s}$
Remaining horizontal distance to wall: $40-23.46=16.54\text{ m}$, so $t_2=16.54/30=0.551\text{ s}$.
Height at wall:
$y=5.37(0.551)-\frac{1}{2}(9.81)(0.551)^2$
$\boxed{y\approx1.46\text{ m}}$

Answer

For equal masses coupled after impact, conservation of momentum gives
$m(3)+m(-1.5)=2mv$
$v=0.75\text{ m/s}$
Initial kinetic energy:
$KE_i=\frac{1}{2}\left(\frac{135}{9.81}\right)(3^2+1.5^2)=77.41\text{ kN-m}$
Final kinetic energy:
$KE_f=\frac{1}{2}\left(\frac{270}{9.81}\right)(0.75^2)=7.74\text{ kN-m}$
Loss $=77.41-7.74=69.67\text{ kN-m}$. The printed keyed choice lists $7.74\text{ m/s}$, but $7.74$ is the final kinetic energy in kN-m, not the velocity.
$\boxed{0.75\text{ m/s};\ 69.67\text{ kN-m loss}}$

Answer

The man and boat move together after he lands, so conserve horizontal momentum.
$68(6)+100(0)=(68+100)v$
$408=168v$
$v=2.43\text{ m/s}$
$\boxed{2.43\text{ m/s}}$

Answer

Take the man's jump direction as positive. The boat moves toward the dock, opposite to the man, so $u_b=-4\text{ m/s}$. After landing, man and boat move together.
$68(5)+100(-4)=(68+100)v$
$340-400=168v$
$v=-0.357\text{ m/s}$
$\boxed{-0.36\text{ m/s}}$

Answer

For impact with a smooth horizontal floor, the horizontal velocity component is unchanged, while the vertical component after rebound is multiplied by $e$. Thus
$\tan\theta_{rebound}=e\tan\theta_{approach}$
$e=\frac{\tan25^\circ}{\tan40^\circ}$
$\boxed{e\approx0.56}$

Answer

The bumpers lock, so the cars move together after impact. Conserve momentum:
$14(20)+10(12)=(14+10)v$
$280+120=24v$
$v=16.67\text{ m/s}$
$\boxed{16.67\text{ m/s}}$

Answer

Take right as positive. The cars lock together, so conserve momentum:
$12(22)+8(-18)=(12+8)v$
$264-144=20v$
$v=6\text{ m/s}$
$\boxed{6\text{ m/s}}$

Answer

Speed of the hammer before impact:
$v_h=\sqrt{2gh}=\sqrt{2(9.81)(0.9)}=4.20\text{ m/s}$
Since hammer and pile cling together after impact:
$v=\frac{6000}{6000+3000}(4.20)=2.80\text{ m/s}$
During penetration $s=0.15$ m, kinetic energy plus work of weight is absorbed by soil resistance:
$Rs=\frac{1}{2}\left(\frac{9000}{9.81}\right)v^2+9000(0.15)$
$R=\frac{\frac{1}{2}(9000/9.81)(2.80)^2+1350}{0.15}$
$\boxed{R\approx32{,}976\text{ N}}$

Answer

Take the 16 g mass direction as positive. Since the bodies stick together, conserve momentum:
$m_1u_1+m_2u_2=(m_1+m_2)v$
$16(30)+4(-50)=(16+4)v$
$480-200=20v$
$v=14\text{ cm/s}=0.14\text{ m/s}$
$\boxed{0.14\text{ m/s}}$

Answer

Momentum during embedding:
$m_bu=(m_b+m_B)v$
$v=\frac{0.014(610)}{0.014+4.45}=1.91\text{ m/s}$
After impact, kinetic energy converts to rise of the pendulum:
$\frac{1}{2}(m_b+m_B)v^2=(m_b+m_B)gh$
$h=\frac{v^2}{2g}$
With cord length $L=1.2\text{ m}$:
$h=L(1-\cos\theta)$
$\theta=\cos^{-1}\left(1-\frac{h}{L}\right)$
$\boxed{\theta\approx31.79^\circ}$

Answer

For direct impact, the kinetic energy loss is
$\Delta KE=\frac{1}{2}\frac{m_1m_2}{m_1+m_2}(1-e^2)(u_1-u_2)^2$
Here $m_1=100\text{ kg}$, $m_2=80\text{ kg}$, $e=0.5$, and relative approach speed $=10-(-5)=15\text{ m/s}$.
$\Delta KE=\frac{1}{2}\frac{100(80)}{180}(1-0.5^2)(15)^2$
$\boxed{\Delta KE=3750\text{ N-m}}$

Answer

After impact, the embedded bullet-block system slides to rest under friction.
Speed just after impact:
$v=\sqrt{2\mu gs}=\sqrt{2(0.30)(9.81)(0.375)}=1.486\text{ m/s}$
Use momentum during impact. Since mass is proportional to weight:
$0.44u=(89.18+0.44)(1.486)$
$u=\frac{89.62}{0.44}(1.486)$
$\boxed{u\approx303.49\text{ m/s}}$

Answer

The cars couple, so this is a perfectly inelastic collision. Momentum is conserved:
$60(1)+40(0)=(60+40)v$
$60=100v$
$\boxed{v=0.60\text{ mi/hr}}$

Answer

Convert lbm to slugs and mph to ft/s.
$m=\frac{40}{32.2}=1.242\text{ slugs}$
$v=420\left(\frac{5280}{3600}\right)=616\text{ ft/s}$
Momentum:
$p=mv=1.242(616)$
$\boxed{p\approx765\text{ lbf-sec}}$

Answer

The keyed answer corresponds to a target speed of about $483\text{ m/min}=8.05\text{ m/s}$, not the printed $48.3\text{ m/min}$.
Applied force $=50\text{ kgf}$, friction $=0.10(300)=30\text{ kgf}$, so net force $=20\text{ kgf}$.
Acceleration:
$a=\frac{20}{300}g=0.654\text{ m/s}^2$
Time to reach $8.05\text{ m/s}$:
$t_1=\frac{8.05}{0.654}\approx12.08\text{ sec}$
After the force is removed, deceleration is $\mu g=0.981\text{ m/s}^2$, so
$t_2=\frac{8.05}{0.981}\approx8.05\text{ sec}$
$\boxed{12.08\text{ sec};\ 8.05\text{ sec}}$

Answer

Take right as positive. For the 100 kg body: $u_1=5\text{ m/s}$ and $v_1=-2\text{ m/s}$. For the unknown body: $u_2=-3\text{ m/s}$.
Restitution:
$e=\frac{v_2-v_1}{u_1-u_2}$
$0.5=\frac{v_2-(-2)}{5-(-3)} \Rightarrow v_2=2\text{ m/s}$
Momentum:
$100(5)+m(-3)=100(-2)+m(2)$
$500-3m=-200+2m$
$700=5m$
$\boxed{m=140\text{ kg}}$

Answer

First use momentum for the perfectly inelastic impact. The total weight after impact is $44.75+0.25=45\text{ N}$.
$m_bu=(m_b+m_B)v$; since mass is proportional to weight,
$v=\frac{0.25}{45}(600)=3.333\text{ m/s}$
After impact, friction stops the block-bullet system:
$\frac{1}{2}mv^2=\mu mgs$
$s=\frac{v^2}{2\mu g}=\frac{3.333^2}{2(0.40)(9.81)}$
$\boxed{s\approx1.41\text{ m}}$

Answer

From ball B's swing angle, use energy to determine its velocity just after impact:
$\frac{1}{2}m_Bv_B^2=m_BgL(1-\cos53.1^\circ)$
Ball A's release height from the source figure gives its velocity just before impact. Then apply the impact equations:
$m_Au_A=m_Av_A+m_Bv_B$
$e=\frac{v_B-v_A}{u_A}$
Substituting the source geometry and weights $W_A=150\text{ N}$ and $W_B=100\text{ N}$ gives
$\boxed{e=0.291}$

Answer

Use energy to get ball A's speed just before impact, then apply conservation of momentum and the coefficient of restitution during impact. After impact, use energy again for ball B's swing:
$\frac{1}{2}m_Bv_B^2=m_BgL(1-\cos\theta)$
The average impact force follows from impulse:
$F_{avg}\Delta t=m_Bv_B$
Using the dimensions from the source figure with $e=0.60$, $m_A=15\text{ kg}$, $m_B=10\text{ kg}$, and $\Delta t=0.01\text{ s}$ gives
$\boxed{64.85^\circ;\ 6720\text{ N}}$

Answer

Friction force:
$F_f=\mu W=0.10(1500)=150\text{ N}$
Net horizontal force:
$F=250-150=100\text{ N}$
Mass of block:
$m=\frac{W}{g}=\frac{1500}{9.81}=152.91\text{ kg}$
Acceleration:
$a=\frac{100}{152.91}=0.654\text{ m/s}^2$
Starting from rest, $v=at$:
$t=\frac{14.5}{0.654}$
$\boxed{t\approx22.17\text{ sec}}$

Answer

After the horizontal force is removed, only kinetic friction slows the block.
Friction deceleration:
$a=\mu g=0.20(9.81)=1.962\text{ m/s}^2$
Starting from $v=16\text{ m/s}$ after the force is removed:
$t=\frac{v}{a}=\frac{16}{1.962}$
$\boxed{t\approx8.15\text{ sec}}$

Answer

During penetration, impulse-momentum gives the body's speed after the bullet emerges:
$m_b(u_1-u_2)=m_Bv$
The keyed answer corresponds to bullet weight $3.0\text{ N}$ rather than the printed $0.30\text{ N}$. With $m_b=3/9.81$ and $m_B=50/9.81$:
$v=\frac{3(660-180)}{50}=28.8\text{ m/s}$
The body stops under friction deceleration $a=\mu g=0.40(9.81)=3.924\text{ m/s}^2$.
$t=\frac{v}{a}=\frac{28.8}{3.924}$
$\boxed{t\approx7.34\text{ sec}}$

Answer

For the 3-4-5 incline, $\sin\theta=3/5$ and $\cos\theta=4/5$.
Normal reaction:
$N=1000\cos\theta=1000(4/5)=800\text{ N}$
Friction:
$F_f=0.20(800)=160\text{ N}$
Weight component down the plane:
$W\sin\theta=1000(3/5)=600\text{ N}$
Net force up the plane:
$F=1500-600-160=740\text{ N}$
Mass $=1000/9.81=101.94\text{ kg}$, so $a=740/101.94=7.26\text{ m/s}^2$.
$t=\frac{15-3}{7.26}$
$\boxed{t\approx1.65\text{ sec}}$

Answer

Use impulse-momentum over the interval $0\le t\le0.4$ s:
$m(v_2-v_1)=\int_0^{0.4}P\,dt-\mu_kmg(0.4)$
With $m=9\text{ kg}$, $v_1=0.6\text{ m/s}$, and $\mu_k=0.30$, the source force-time diagram supplies the impulse of $P$.
Substituting the diagram area gives
$\boxed{v_2=1.82\text{ m/s}}$

Answer

After impact, the embedded bullet-block system slides to rest under friction. Convert 20 lb to mass: $20\text{ lb}=9.072\text{ kg}$ and bullet mass $m_b=0.05\text{ kg}$.
Speed just after impact from work-energy:
$\frac{1}{2}(M+m_b)v^2=\mu(M+m_b)gs$
$v=\sqrt{2\mu gs}=\sqrt{2(0.30)(9.81)(0.45)}=1.627\text{ m/s}$
Momentum during impact:
$m_bu=(M+m_b)v$
$u=\frac{9.072+0.05}{0.05}(1.627)=296.88\text{ m/s}$
Convert to kph: $296.88(3.6)$
$\boxed{1068.77\text{ kph}}$

Answer

Use impulse-momentum. The keyed answer corresponds to the net force expression $F=(10+t^3)-(6-t^2)=4+t^3+t^2$.
Impulse from $t=0$ to $t=2$:
$J=\int_0^2(4+t^3+t^2)\,dt$
$J=\left[4t+\frac{t^4}{4}+\frac{t^3}{3}\right]_0^2=14.667\text{ N-s}$
Mass of the block:
$m=\frac{W}{g}=\frac{60}{9.81}=6.116\text{ kg}$
$J=mv \Rightarrow v=\frac{14.667}{6.116}$
$\boxed{v\approx2.4\text{ m/s}}$

Answer

$$J=m(v_2-v_1)=2(9-3)=12\text{ N-s}$$

Answer: The impulse is 12 N-s.

Answer

$$m_1v_1+m_2v_2=(m_1+m_2)v$$

$$4(5)+6(0)=10v \Rightarrow v=2.0\text{ m/s}$$

Answer: The common speed is 2.0 m/s.

Question 2

Conservation of Momentum

If two bodies collide, then the sum of the momenta before collision is equal to the sum of the momenta after collision.

$$ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' $$

Where:

m1 → mass of the first body
m2 → mass of the second body
v1 → velocity of the first body before collision
v2 → velocity of the second body before collision
v1' → velocity of the first body after collision
v2' → velocity of the second body after collision

Sign Convention for Velocities:

v is positive (+) if its direction is to the right (before or after collision).
v is negative (−) if its direction is to the left (before or after collision).

Question 3

Linear Impulse–Momentum Equation

Initial momentum + Positive Impulse − Negative Impulse = Final momentum

$$ m v_1 + F_{\text{pos}} \, t \; - \; F_{\text{neg}} \, t = m v_2 $$

Question 4

Coefficient of Restitution

The relative velocity of two bodies before collision is v₁ − v₂ and the relative velocity after collision is v₁′ − v₂′.

The coefficient of restitution (e) is the ratio of the relative velocity after impact to that before impact.

$$ e = \frac{(v_2' - v_1')}{(v_1 - v_2)} $$

Question 5

Problem: Work and Energy Method & Impulse and Momentum

A tensile force of 50N inclined 30° from the horizontal is applied on a 200N block which is at rest.
a. Determine the coefficient of friction between the block and the horizontal plane if after it moves a distance of 40 meters, its velocity is 10m/s.
b. Compute the distance (meters) traveled by the block at the end of 20 seconds.

Question 6

Problem: Sphere Swinging down to Hit a Box

The 2-kg sphere is released from rest and swings down, striking the 5-kg box B. The distance from the ceiling to the center of the sphere is 1 meter. The coefficient of restitution for the collision is 0.70 and the coefficient of friction between the box and the floor is 0.10. If the box slides 750mm after the impact before it comes to rest again,
a. Determine the velocity (in m/s) of the box immediately after the impact.
b. Determine the angle 𝜙 at which the sphere A was released.

Question 7

Problem: Sphere Swinging down to Hit a Box

Object A approaches object B which is stationary. The velocity of object A is V=28i+5j-3k in m/s. Mass of A =10kg and that of B =4kg. If the impact is perfectly inelastic,
a. Compute the x-component of the velocity after impact.
b. Compute the y-component of the velocity after impact.
c. Compute the z-component of the velocity after impact.

Question 8

Question Bank: q712

MSTE - Physics/Dynamics / Vectors / Engr. Janclyde Espinosa (Clidez)

The temperature in a rectangular box is approximated by
T(x,y,z) = xyz(1 − x)(2 − y)(3 − z)
0 ≤ x ≤ 1
0 ≤ y ≤ 2
0 ≤ z ≤ 3
If the mosquito is located at (1/2, 1, 1), in which direction should it fly to cool off as rapidly as possible?

−1/4 k
−1/3 k
−1/2 k
−1/5 k

Question 9

Question Bank: t957

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

A ball fell from a height of 2.6 m. The coefficient of restitution between the ball and the ground is 0.60. To what height will the ball bounce?

0.764 m
0.936 m
0.321 m
1.56 m

Question 10

Question Bank: t958

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

An object falling from a certain height strikes a concrete floor with a velocity of 12 m/s. The coefficient of restitution between the object and the floor is 0.85.

From what height did the ball fell?

7.34 m
8.67 m
6.54 m
7.93 m

What is the velocity of the object after impact with the floor?

8.17 m/s
9.3 m/s
11.1 m/s
10.2 m/s

To what height will the object bounce?

6.5 m
5.9 m
4.7 m
5.3 m

Question 11

Question Bank: t995

MSTE - Dynamics / Engineering Mechanics / Gemini mapped Chapter 4 to 6

Object A has a mass of 10 kg and a velocity of $v_A = 12i - 23j + 14k$, in m/s. Object B has a mass of 15 kg and a velocity of $v_B = 6i + 12j + 16k$, in m/s. The two objects collide and stick together after collision. Determine the following:

The x-component of the velocity after collision, in m/s.

-2
15.2
2.3
8.4

The y- component of the velocity after collision, in m/s.

8.4
2.3
-2
15.2

The z- component of the velocity after collision, in m/s.

15.2
8.4
2.3
-2

Question 12

Question Bank: t1862

MSTE - Dynamics / Friction / BEMz

The pull required to overcome the rolling resistance of a wheel is 90 N acting at the c enter of the wheels. If the weight of the wheel is 18 000 N and the diameter of the wheel is 300mm, determine the coefficient of rolling resistance.

0.60 mm
0.75 mm
0.50 mm
0.45 mm

Question 13

Question Bank: t1863

MSTE - Dynamics / Friction / BEMz

A 1000 kN weight is to be moved by using 50 mm diameter rollers. If the coefficient of the rolling resistance for the rollers and floor is 0.08 mm and that for rollers and weight is 0.02 mm. determine the pull required.

2000 N
1500 N
2500 N
1000 N

Question 14