A car with mass 1155 kg, moving at 32 m/s,strikes a(n) 2553 kg car at rest.If the two cars stick together, with whatspeed do they move? answer in m/s.  

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The car moving with 1155kg at 32 m/s has a momentum which is  the product of its mass and velocity = 115kg*32m .

The momentum of the car of 2553 kg at rest (or zero velocity = 2553kg * 0 = 0

After collision, the cars sticks. And move with a speed v which is to be determined. The momentum of the cars after the collison is their sum of masses times velocity v = (1155kg+2553kg)v

Considering the law of conservation of momentum  in collision:

Momentum before collision =momentum after collision

1155*32 = (1155+ 2553)V

or v = 9.9675 m/s. The cars are not elastic bodies and there must be loss of energy resulting in heat sound and deshaping, whereas momentum is conserved even in inelastic cases.

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

In this problem the kinetic energy of the first car before striking the second car will be same as combined the kinetic energy of the two cars when they move together.

We know kinetic energy (k) of a moving object is given by the formula:

k = (mv^2)/2

Given:

Mass of car 1 = 1155 kg = m1

Velocity of car 2 = 32 m/s = vb

Mass of car 2 = 2553 kg = m2

We have to find combined velocity (va) of the to two cars after striking.

Kinetic energy of car 1 before striking (kb) is calculated from the formula of kinetic energy as:

kb = (m1*vb^2)/2 = (1155*32^2)/2 = 1155*512

Similarly kinetic energy of both cars after striking (ka) is given by:

kb = [(m1+m2)*vb^2)]/2 = [(1155+2553)*vb^2)]/2 = 1854*vb^2

We know kb = ka

Therefore: 1155*512 = 1854*vb^2

Therefore: vb^2 = (1155*512)/1854 = 17.8596 m/s

Answer:

Two cars move together with speed of 17.8596 m/s.

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