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3 years ago

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A 1050 kg sports car is moving westbound at 14.0 m/s on a level road when it collides with a 6320 kg truck driving east on the s

ame road at 10.0 m/s. The two vehicles remain locked together after the collision.a) What is the velocity (magnitude) of the two vehicles just after the collision?b) What is the direction of the velocity of the two vehicles just after the collision?c) At what speed should the truck have been moving so that it and car are both stopped in the collision?d) Find the change in kinetic energy of the system of two vehicles for the situations of part a).

1 answer:

goldfiish [28.3K]3 years ago

5 0

Answer:

(a) 6.58 m/s

(b) East direction

(c) 2.33 m/s

(d) -259353 J

Explanation:

From the law of conservation of linear momentum

M1u1+m2u2=(m1+m2)V where V is common velocity, m1 and m2 are masses of objects, u1 and u2 are velocities of m1 and m2 respectively

Making V the subject

$V=\frac {m1u1+m2u2}{m1+m2}$

Let West be positive while East be negative hence

$V=\frac {1050*14+6320*-10}{1050+6320}= -6.58073 m/s\approx -6.58 m/s$

Velocity=6.58 m/s

(b)

Since common velocity is negative, it implies that after collision the vehicles move to the East

(c)

m1u1=m2u2

$u1=\frac {m2u2}{m1}=\frac {1050*14}{6320}= 2.325949367 m/s\approx 2.33 m/s$

(d)

The initial kinetic energy is given by

$KE_i=0.5m1(u1)^{2}+0.5m2(u2)^{2}=0.5(m1(u1)^{2}+ m2(u2)^{2})=0.5(1050*14^{2}+6320*10^{2}}= 418900 J$

The final kinetic energy is given by

$KE_f=0.5(m1+m2)V^{2}=0.5(1050+6320)*6.58^{2}= 159547.2 J$

Since the two vehicles remain locked after collision, this is inelastic collision hence kinetic energy is not conserved

Change in kinetic energy is given by

$\triangle KE=KE_f-KE_i=159547.2 J-418900 J=-259352.766 J\approx -259353 J$

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