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SOLUTION Find the x-components of the initial Epxi = mcarVcar = (1.50 × 10³ kg)(25.0 m/s) = 3.75 x 104 kg • m/s and final total momenta. Epxf = (mcar + mvan)Vf cos 0 = (4.00 × 10³ kg)v, cos 0 Set the initial x-momentum equal to (1) 3.75 x 10ª kg · m/s = (4.00 × 103 kg)vf cos 0 the final x-momentum. Find the y-components of the initial Epiy mvanVvan = (2.50 × 103 kg)(20.0 m/s) = 5.00 × 104 kg · m/s and final total momenta. Epfy = (m car + myan)Vf sin 0 = (4.00 × 103 kg)vf sin 0 Set the initial y-momentum equal to (2) 5.00 x 104 kg · m/s = (4.00 × 103 kg)vf sin 0 the final y-momentum. Divide Equation (2) by Equation (1) 5.00 x 104 kg • m/s 3.75 x 104 kg • m/s tan 0 = = 1.33 and solve for 0. 0 = 53.1° Substitute this angle back into 5.00 x 104 kg • m/s Vf = (4.00 × 103 kg) sin 53.1° 15.6 m/s Equation (2) to find LEARN MORE REMARKS It's also possible to first find the x- and y-components Vfx and of the resultant velocity. The magnitude and direction of the resultant velocity can then be found with the Pythagorean theorem, vf Vv2 + va?, and the inverse tangent function 0 = tan-1 (va/VA). Setting up this alternate approach is a simple matter of substituting v = vf cos 0 and vâ = vf sin 0 in Equations (1) and (2). = VVA Vfy QUESTION If the car and van had identical mass and speed, what would the resultant angle have been? 50

SOLUTION Find the x-components of the initial Epxi = mcarVcar = (1.50 × 10³ kg)(25.0 m/s) = 3.75 x 104 kg • m/s and final total momenta. Epxf = (mcar + mvan)Vf cos 0 = (4.00 × 10³ kg)v, cos 0 Set the initial x-momentum equal to (1) 3.75 x 10ª kg · m/s = (4.00 × 103 kg)vf cos 0 the final x-momentum. Find the y-components of the initial Epiy mvanVvan = (2.50 × 103 kg)(20.0 m/s) = 5.00 × 104 kg · m/s and final total momenta. Epfy = (m car + myan)Vf sin 0 = (4.00 × 103 kg)vf sin 0 Set the initial y-momentum equal to (2) 5.00 x 104 kg · m/s = (4.00 × 103 kg)vf sin 0 the final y-momentum. Divide Equation (2) by Equation (1) 5.00 x 104 kg • m/s 3.75 x 104 kg • m/s tan 0 = = 1.33 and solve for 0. 0 = 53.1° Substitute this angle back into 5.00 x 104 kg • m/s Vf = (4.00 × 103 kg) sin 53.1° 15.6 m/s Equation (2) to find LEARN MORE REMARKS It's also possible to first find the x- and y-components Vfx and of the resultant velocity. The magnitude and direction of the resultant velocity can then be found with the Pythagorean theorem, vf Vv2 + va?, and the inverse tangent function 0 = tan-1 (va/VA). Setting up this alternate approach is a simple matter of substituting v = vf cos 0 and vâ = vf sin 0 in Equations (1) and (2). = VVA Vfy QUESTION If the car and van had identical mass and speed, what would the resultant angle have been? 50

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter6: Momentum, Impulse, And Collisions
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Problem 46P: A space probe, initially at rest, undergoes an internal mechanical malfunction and breaks into three...
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SOLUTION Find the x-components of the initial Epxi = mcarVcar = (1.50 × 103 kg)(25.0 m/s) and final total momenta. = 3.75 x 104 kg • m/s (mcar + mvan) vf cos 0 = (4.00 × 103 kg)v, cos 0 Ep.f Set the initial x-momentum equal to (1) 3.75 x 104 kg · m/s = (4.00 × 103 kg)vf cos 0 the final x-momentum. Find the y-components of the initial Epiy (2.50 × 10³ kg)(20.0 m/s) = mvanVvan = 5.00 x 104 kg • m/s and final total momenta. Efy (mcar + mvan)Vf sin 0 = (4.00 × 103 kg)vf sin 0 Set the initial y-momentum equal to the final y-momentum. (2) 5.00 x 104 kg · m/s = (4.00 × 103 kg)vf sin 0 Divide Equation (2) by Equation (1) 5.00 x 104 kg • m/s 3.75 x 104 kg • m/s tan 0 = = 1.33 and solve for 0. 0 = 53.1° Substitute this angle back into 5.00 x 104 kg • m/s = 15.6 m/s Equation (2) to find vf. (4.00 x 103 kg) sin 53.1° LEARN MORE REMARKS It's also possible to first find the x- and y-components v and VA of the resultant velocity. The magnitude and direction of the resultant velocity can then be found with the Pythagorean theorem, vf = Vv2 + v62, and the inverse tangent function 0 = tan-1 (V/Vfr). Setting up this alternate approach is a simple matter of substituting v = vf cos 0 and v, = vf sin 0 in Equations (1) and (2). QUESTION If the car and van had identical mass and speed, what would the resultant angle have been? 50
Transcribed Image Text:SOLUTION Find the x-components of the initial Epxi = mcarVcar = (1.50 × 103 kg)(25.0 m/s) and final total momenta. = 3.75 x 104 kg • m/s (mcar + mvan) vf cos 0 = (4.00 × 103 kg)v, cos 0 Ep.f Set the initial x-momentum equal to (1) 3.75 x 104 kg · m/s = (4.00 × 103 kg)vf cos 0 the final x-momentum. Find the y-components of the initial Epiy (2.50 × 10³ kg)(20.0 m/s) = mvanVvan = 5.00 x 104 kg • m/s and final total momenta. Efy (mcar + mvan)Vf sin 0 = (4.00 × 103 kg)vf sin 0 Set the initial y-momentum equal to the final y-momentum. (2) 5.00 x 104 kg · m/s = (4.00 × 103 kg)vf sin 0 Divide Equation (2) by Equation (1) 5.00 x 104 kg • m/s 3.75 x 104 kg • m/s tan 0 = = 1.33 and solve for 0. 0 = 53.1° Substitute this angle back into 5.00 x 104 kg • m/s = 15.6 m/s Equation (2) to find vf. (4.00 x 103 kg) sin 53.1° LEARN MORE REMARKS It's also possible to first find the x- and y-components v and VA of the resultant velocity. The magnitude and direction of the resultant velocity can then be found with the Pythagorean theorem, vf = Vv2 + v62, and the inverse tangent function 0 = tan-1 (V/Vfr). Setting up this alternate approach is a simple matter of substituting v = vf cos 0 and v, = vf sin 0 in Equations (1) and (2). QUESTION If the car and van had identical mass and speed, what would the resultant angle have been? 50
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