b) Derive an expression for the direction and approximate magnitude of the force

Question

b) Derive an expression for the direction and approximate magnitude of the force experienced by m1 during the collision.

c) Derive an expression for the change in kinetic energy of m1 during the collision and the average power delivered to m1 during the collision.

When answering c) also express the result in terms of kg of TNT. Each kg of TNT contains 7.5MJ of chemical potential energy.

Two cars [m1, m2] are initially moving with constant velocities [v1(+x)),v2(-x) m/s] on a horizontal surface [ ] and interact in a perfectly inelastic collision with duration [ t = 0.45s and evaluate each of your expressions. When answering c) also express the result in terms of kg of TNT. Each kg of TNT contains 7.5MJ of chemical potential energy. r. b) Derive an expression for the direction and approximate magnitude of the force experienced by m1 during the collision. c) Derive an expression for the change in kinetic energy of m1 during the collision and the average power delivered to m1 during the collision. d) Use data for a Volvo [m1 = 1350kg, v1 = 23m/s] and a Mercedes [m2 = 1850kg, v2 = 29m/s], and the road [1/2 s = 0.80, I 1/2 k = 0.65] with r m] from the point where the collision took place. Assume all forces act in a plane parallel to the road surface and the system center of mass does not displace significantly during the collision (v1, v2 >> Vcom). a) Derive an expression for the magnitude and direction of t s]. Afterwards the cars skid to a halt, stopping [

Explanation / Answer

a)m1v1-m2v2=(m1+m2)v

v=(m1v1-m2v2)/(m1+m2)

Conserving energy,

u(m1+m2)g(delta r)=0.5*(m1+m2)*((m1v1-m2v2)/(m1+m2))^2

delta r=0.5*((m1v1-m2v2)/(m1+m2))^2/ug

b)Force=change in momentum/time=m1v1-(m1^2v1-m1m2v2)/Delta t(m1+m2)= (m1^2v1+m1v1m2-m1^2v1+m1m2v2)/(m1+m2)=m1m2(v1+v2)/(delta t(m1+m2))

c)Change in KE=0.5*m1*((m1v1-m2v2)/(m1+m2))^2-0.5m1v1^2

Power delivered=Change in KE/delta t=(0.5*m1*((m1v1-m2v2)/(m1+m2))^2-0.5m1v1^2)/delta t

d)a)r=0.5*((m1v1-m2v2)/(m1+m2))^2/(ug)=3.91m

b)F=m1m2(v1+v2)/(delta t(m1+m2))=1350*1850*(23+29)/(0.45*(1350+1850))=90188N

c)0.5*m1*((m1v1-m2v2)/(m1+m2))^2-0.5m1v1^2=0.5*1350*((1350*23-1850*29)/(1350+1850))^2-0.5*1350*23^2=

323407J=0.043kg TNT

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