physics motion questions, 1st year physics

8/10/2019 Physics motion questions, 1st year physics

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Problem Class Week 3: Problem 3.1

Blocks of wood of masses 1.0, 2.0 and 3.0 kg are lined up on a SMOOTH tableas shown. A rightwards-pointing 12 Newton force is applied to the left-mostblock. What force does the middle block exert on the rightmost one?

Solution: All forces on the rightmost block are shown in black. In the

vertical direction the weight force W=3g is cancelled by the normalreaction N exerted by the table, leaving a total vertical force of zero.

Since there is no friction, the only other force on the 3 kg block is the

contact force C as shown: C is a reaction force due to the presence of

the 2 kg block, and it is the force that we are asked to find.

Newton II gives this force as C = ma = 3a, where a is the acceleration

of the 3rd

block.So we can find Cprovided that we can find the acceleration a.

Now because the external rightwards-directed 12 N force is applied to

the leftmost block, all the blocks are pushed to the right together, so

they all experience a common acceleration a.

The only horizontal EXTERNAL force on the composite body (all

three blocks, total mass 6 kg) is the 12N force. (The force C shown isan INTERNAL force. The INTERNAL forces of the composite body

cancel in action-reaction pairs.) Thus Newton II for the 3 blocks

together reads

F = Ma

12 = 6a

a = 2 m/s2

Then from above C = 3a = (3)(2) = 6 Newton

3 kg2 kg1 kg

N=3g

W=3g

C

12

newton


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Problem Class Wk 3 : Problem 3.2

A Navy jet of mass 10,000 kg lands on an aircraft carrier and snags a cable toslow it down. The cable is attached to a spring of stiffness constant 40,000 N/m.

If the spring stretches 25 m to stop the plane, what was the landing speed of theplane? 25 m

v0 = ?

Just arrived Stopped

The potential energy of the spring is kx2 and the kinetic energy of the plane tied

to the spring is mv2 (mass & KE of spring are negligible).

As the plane is slowed down, its kinetic energy is converted to potential energy

of the stretched spring. The total energy is conserved (constant) during this

process, under the assumption of negligible friction.Conservation of energy: (KE + PE)before = (KE + PE)after

(mv2 + kx2)before = (mv2 + kx2)after

mv02 + k(0)2 = m(0)2 + k(25)2

v02 = (25)2k/m

v0 = 25(k/m)1/2 = 25(40000/10000)1/2 = (25)(2) = 50Landing velocity is 50 m/s


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Problem Class Week 3 : Prob 3.3

A simple model of an automobile suspension consists of a mass attached to a

spring. If the mass is 1900 kg and the stiffness constant is 26 kN/m,

(i) with what period will the car undergo simple harmonic motion?

(ii) If the amplitude of the SHM is 5 cm, what is the max vertical speed of the car(assuming shock absorbers disconnected so we have pure spring motion)?

v = 0

|v| = vmax

A

x

t

METHOD 1: ENERGY CONSERVATION(KE + PE)

x=A= (KE + PE)

x=0m02 +kA2 =mv

max2 +k02

(k/m)A2 = vmax

2

vmax

= (k/m)1/2A

= (26000/1900)1/2

(0.05)= 0.18 m/s

This is a plug-in problem (see lec slide 70) :T = 2(m/k)1/2 = 2(1900/26000)1/2 = 1.7 sec.

Method 1: conservation of energy. The energymv2 +kx2 =E is constant during the motion. Bothmv2 andkx2 are never negative, so v2 is maximum when x2 is minimum = i,e, when x=0. By the same

reasoning, v is zero when x is a max - i.e. when x = A, the amplitude. (the same conclusions follow fromlooking at the sketch of x vs t in simple harmonic motion below.)

Method 2: equation for displacement vs time in SHM

x = Acos(t+), v = -Asin(t+). Max of this last expression is -A(-1) = A,

occuring when sin(t+) = -1. Since =(k/m) this givesvmax = (k/m)1/2A same as in method 1 above,


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A railway carriage of mass 3 tonnes travelling at 3 m/s comes up behind a

carriage of mass 6 tonnes that is moving at 1 m/s in the same direction. Theycouple together upon impact.

(a) What is the speed of the combined carriages straight after the collision?

3000 kg 6000 kg 9000 kg

BEFORE AFTER

3 m/s 1 m/sV

METHOD: CONSERVATION OF MOMEMTUM(Reasoning: This is a rapid collision with strong internal impulsive forces, so for the brief duration of the collision we can ignoreexternal forces such as friction. Then total momentum is conserved,)

Total momentum before = total momentum after (with p = mv in genera(3000)(3) + (6000)(1) = (9000)V

V = (9+6)/9 = 5/3 = 1.66 m/sec

(b) If they are on a long level section of track, what is their velocity a long time afterthe collision?Answer: V = 0. Reasoning: During the brief collision, non-impulsive forces such as friction could be ignored compared withimpulsive internal forces. However over a longer time friction will certainly slow the carriages down and bring them to restNote that the answer in part (a) above is the velocity immediately afterthe coupling collision.

Problem Class Week 3 : Problem 3.4

physics motion questions, 1st year physics

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