physics problem

[ashkanOo]

okey

another question

come on guys

just answer

[klausmeister]

A 1400 kg car, heading north and moving at 35 miles per hour collides in a perfectly inelastic collision with a 4000 kg truck going East at 20 miles per hour.

a.
What is the speed and direction of the wrecked vehicles just after collision?
b.
What percentage of the total mechanical energy is lost from the collision?

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This problem is concerned with conservation of linear momentum. The collision is said to be inelastic, that is the two cars stick together right after the collision. In this case, since both cars are stuck together after the event, their bodies are also deformed, so total energy before the collision has been changed/transformed due to deformation and heat.

Also, momentum is a vector quantity. This means we have to add vectors in order to find the resultant vector which can be done graphically as well as algebraically.

a)

Total momentum before the collision is equal to total momentum after the collision.
Pc,i + Pt,i = mc * vc,i + mt * vt,i = m(c+t) *vr,f = P(c+t),f

Here, P, and v are in bold letters to signify them as vectors. The subscripts c, and t stand for car and truck, respectively; (c+t) designates the state when car and truck are stuck together. m is the mass of the car or truck. Subscripts i and f are short for initial and final state, respectively.

Since we adhere to SI units, we convert miles to kilometers:

35 miles = 56.33 km and 20 miles = 32.19 km.

We first calculate momenta as scalar:

Pc = mc * vc = 1400 kg * 56.33 km/hr =78,862.00 kg*km/hr

Pt = mt * vt = 4000 kg * 32.19 km/hr = 128,760.00 kg*km/hr

m(c+t) = 5400 kg

Since the car is travelling due north and the truck due east, after the crash, the resulting momentum vector, P(c+t),f can be obtained by adding momentum vectors of car and truck using Pythagoras theorem:

Pc,i +Pt,i = P(c+t),f

Since

= P.P = P*P*Cos(0). That is P is pointing in the same direction, therefore the angle is 0.

P(c+t),f =

From above equation we have: P(c+t),f = m(c+t)*vr,f

So, the speed of the car and truck combined right after the collision is: vr,f =

vr,f = 27.96 km/hr

The direction of the resultant vector, i.e. that of combined car and truck, is obtained by : tan a = Pt/Pc

a = arctan (Pt/Pc)

a= 58.51 degrees

So, the combined wreck of car and truck after the collision will travel at a speed of 27.96 km/hr due east of north.

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b)

To find the lost energy due to collision, we subtract the initial total energy from the final total energy:

Total energy before the collision, KE tot,i:

=1/2 *1400 kg * 56.33^{2}(km/hr)^2 + 1/2 * 4000 kg* 32.19 ^{2}(km/hr)^2

KE tot,i = 4293540.43 kg*( km/hr)^2

Total energy after the collision, KE tot, f:

=1/2 * 5400 kg * 27.96 ^2 (km/h)^2

KE tot,f = 2110756.32 kg*( km/hr)^2

So, the percentage of lost energy is:

=

*100 %

=0.508 * 100 %

=50.8 %

[ashkanOo]

A 0.25Kg block oscillates on the end of the spring with a spring constant of 200 Newton/meter. If the oscillation is started by elongating the spring 0.15meter and giving the block a speed of 3.0 meter/second, then what is the maximum speed of the block?
A. 0.13 m/s
B. 0.18 m/s
C. 3.7 m/s
D. 5.2 m/s