Question

An Escort and a Camaro traveling at right angles collide and stick together. The Escort has a mass of 1100 kg and a speed of 25 km/h in the positive x direction before the collision. The Camaro has a mass of 1600 kg and was traveling in the positive y direction. After the collision, the two move off at an angle of 49 o to the x axis. What was the speed of the Camaro?(in km/h)

Answer #1

Given that

The Escort has a mass of (m1) = 1100 kg

The speed of the escort is(u1) = 25 km/h in the positive x direction before the collision.

The Camaro has a mass of (m2) =1600 kg and was traveling in the positive y direction.

After the collision, the two move off at an angle of (theta) = 49degrees to the x axis.

Now from the conservation of momentum

m1u1i+m1u2j=(m1+m2)(v1i+v2j)

Now comparing the equation we get

1100kg*25i+1600*vj =(1100+1600)(vxi+vyj)

1100*25 =2700*vx

v_{x} =10.185km/h

We also can write v_{x} =vcos49

Then v =vx/cos49 =10.185km/h/cos49 =15.524m/s

Now the vertical component of velocity is given by

Vy =vsin49 =(15.524)sin49 =11.716km/h

The speed of the camaro is (v) =21.901km/h

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