Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 7, Problem 69P
To determine

The speed and direction of the tangled mess of metal.

Expert Solution & Answer
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Answer to Problem 69P

The speed and direction of the car is 6.0m/s at 21° S of E.

Explanation of Solution

The figure 1 shows the final position of cars moved together into the fourth quadrant.

Physics, Chapter 7, Problem 69P , additional homework tip  1

The collision betweenthe two cars is perfectly inelastic, so the final velocities of the cars are identical.

Write the equation for initial momentum of the object along x axis.

pix=m1v1ix+m2v2ix (I)

Here, pix is the initial momentum of the car before collsion along x direction, m1 is the mass of the car, v1ix is the initial velocity of the car along north and east direction, Physics, Chapter 7, Problem 69P , additional homework tip  2 is the mass of the smaller car, and v2ix is the initial velocity of the smaller car along north and east direction.

Write the equation for final momentum of the object along x axis.

pfx=(m1+m2)vfx (II)

Here, pfx is the final momentum of the car during collsion along x direction and vfx is the final velocity of the car.

Write the equation for initial momentum of the object aong y axis.

piy=m1v1iy+m2v2iy (III)

Here, piy is the initial momentum of the car after collsion along y direction, v1iy is the initial velocity of the car along north direction, and v2iy is the initial velocity of the smaller car along north direction.

Write the equation for final momentum of the object along y axis.

pfy=(m1+m2)vfy (IV)

Here, pfy is the final momentum of the car along y direction and vfy is the final velocity of the car.

Rewrite the equation (I) by substituting 0 for v2ix.

pix=m1v1ix (V)

Rearrange the equation (II) and (V) for conservation of momentum to solve for vfx.

pix=pfxm1v1ix=(m1+m2)vfxvfx=m1(m1+m2)v1icosθ (VI)

Rearrange the equation (III) and (IV) for conservation of momentum to solve for vfy.

piy=pfym1v1iy+m2v2iy=(m1+m2)vfyvfy=m1v1isinθ+m2v2iy(m1+m2) (VII)

Write the equation for final speed of the cars.

vf=vfx2+vfy2 (VIII)

Write the equation for direction of the cars.

θ=tan1(vfyvfx)                                                                         (IX)

Conclusion:

Substitute 1700kg for m1, 1300kg for m2, 45° for θ, and 14m/s for v1i in equation (VI).

vfx=1700kg(1700kg+1300kg)(14m/s)cos45°=1700kg3000kg(9.9m/s)=5.61m/s

Substitute 1700kg for m1, 1300kg for m2, 45° for θ, 18m/s for v2iy, and 14m/s for v1i in equation (VII).

vfy=(1700kg)(14m/s)sin45°+(1300kg)(18m/s)(1700kg+1300kg)=16829kgm/s23400kgm/s3000kg=2.19m/s

Substitute 2.19m/s for vfy and 5.61m/s for vfx in equation (VIII).

vf=(5.61m/s)2+(2.19m/s)2=6.0m/s

Substitute 2.19m/s for vfy and 5.61m/s for vfx in equation (IX).

θ=tan1(2.19m/s5.61m/s)=tan1(0.390)=21.3°=21°

Therefore, the speed and direction of the car is 6.0m/s at 21° S of E.

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Chapter 7 Solutions

Physics

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