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15–5.

A hockey puck is traveling to the left with a velocity of v1 = 10 m>s when it is struck by a hockey stick and given a velocity of v2 = 20 m>s as shown. Determine the magnitude of the net impulse exerted by the hockey stick on the puck. The puck has a mass of 0.2 kg.

Solutionv1 = {-10i} m>s

v2 = {20 cos 40°i + 20 sin 40°j} m>s

I = m∆v = (0. 2) {[20 cos 40° - (-10)]i + 20 sin 40°j}

= {5.0642i + 2.5712j} kg # m>s

I = 2(5.0642)2 + (2.5712)2

= 5.6795 = 5.68 kg # m>s Ans.

v1 � 10 m/s

v2 � 20 m/s

40�

Ans:I = 5.68 N # s

479

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15–6.

A train consists of a 50-Mg engine and three cars, eachhaving a mass of 30 Mg. If it takes 80 s for the train toincrease its speed uniformly to 40 , starting from rest,determine the force T developed at the coupling betweenthe engine E and the first car A. The wheels of the engineprovide a resultant frictional tractive force F which givesthe train forward motion, whereas the car wheels roll freely.Also, determine F acting on the engine wheels.

km>h

F

vEA

SOLUTION

Entire train:

Ans.

Three cars:

Ans.0 + T(80) = 3(30) A103 B(11.11) T = 12.5 kN

a :+ b m(vx)1 + ©L

Fx dt = m(vx)2

F = 19.4 kN

0 + F(80) = [50 + 3(30)] A103 B(11.11)

a :+ b m(vx)1 + ©L

Fx dt = m(vx)2

(yx)2 = 40 km>h = 11.11 m>s

Ans:F = 19.4 kNT = 12.5 kN

483

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15–10.

30�

PThe 50-kg crate is pulled by the constant force P. If the cratestarts from rest and achieves a speed of in 5 s, deter-mine the magnitude of P. The coefficient of kinetic frictionbetween the crate and the ground is .mk = 0.2

10 m>s

SOLUTIONImpulse and Momentum Diagram: The frictional force acting on the crate is

.

Principle of Impulse and Momentum:

(1)

(2)

Solving Eqs. (1) and (2), yields

Ans.P = 205 N

N = 387.97 N

4.3301P - N = 500

50(0) + P(5) cos 30° - 0.2N(5) = 50(10)

m(v1)x + ©L

t2

t1

Fx dt = m(v2)x(:+ )

N = 490.5 - 0.5P

0 + N(5) + P(5) sin 30° - 50(9.81)(5) = 0

m(v1)y + ©L

t2

t1

Fy dt = m(v2)y(+ c)

Ff = mkN = 0.2N

Ans:P = 205 N

487

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15–14.

A tankcar has a mass of 20 Mg and is freely rolling to the right with a speed of 0.75 m>s. If it strikes the barrier, determine the horizontal impulse needed to stop the car if the spring in the bumper B has a stiffness (a) k S ∞ (bumper is rigid), and (b) k = 15 kN>m.

Solution

a) b) ( S+ ) mv1 + ΣLF dt = mv2

20(103)(0.75) - LF dt = 0

LF dt = 15 kN # s Ans.

The impulse is the same for both cases. For the spring having a stiffness k = 15 kN>m, the impulse is applied over a longer period of time than for k S ∞ .

v � 0.75 m/s

k

B

Ans:I = 15 kN # s in both cases.

488

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15–15.

The motor, M, pulls on the cable with a force F = (10t2 + 300) N, where t is in seconds. If the 100 kg crate is originally at rest at t = 0, determine its speed when t = 4 s. Neglect the mass of the cable and pulleys. Hint: First find the time needed to begin lifting the crate.

SolutionPrinciple of Impulse and Momentum. The crate will only move when 3(10t2 + 300) = 100(9.81). Thus, this instant is t = 1.6432 s. Referring to the FBD of the crate, Fig. a,

( + c ) m(vy)1 + ΣLt2

t1

Fy dt = m(vy)2

0 + L4 s

1.6432 s3(10t2 + 300) dt - 100(9.81)(4 - 1.6432) = 100v

3 a10t3

3+ 300tb `

4 s

1.6432 s- 2312.05 = 100v

v = 4.047 m>s = 4.05 m>s c Ans.

M

Ans:v = 4.05 m>s