Question

PLEASEEEE HEEEEELP!!!!!

Answer 1

Before:

`p_(truck)=16400\ kg.m/s`

`p_(car)=10000\ kg.m/s`

After:

`p_(truck)=8000\ kg.m/s`

`p_(car)=8400\ kg.m/s`

`v_(fcar)=8.4\ m/s`

`F=9333.33 \ Nw`

Step-by-step explanation:

Conservation of Momentum

Two objects of masses m1 and m2 moving at speeds v1o and v2o respectively have a total momentum of

`p_1=m_1v_(1o)+m_2v_(2o)`

After the collision, they have speeds of v1f and v2f and the total momentum is

`p_2=m_1v_(1f)+m_2v_(2f)`

Impulse J is defined as

`J=F.t`

Where F is the average impact force and t is the time it lasted

Also, the impulse is equal to the change of momentum

`J=\Delta p`

As the total momentum is conserved:

`p_1=p_2`

`m_1v_(1o)+m_2v_(2o)=m_1v_(1f)+m_2v_(2f)`

We can compute the speed of the second object by solving the above equation for v2f

`\displaystyle v_(2f)=( m_1v_(1o)+m_2v_(2o) -m_1v_(1f) )/( m_2 )`

The given data is

`m_1=2000\ kg`

`m_2=1000\ kg\\v_(1o)=8.2\ m/s\\v_(2o)=0\ m/s\\v_(1f)=4\ m/s`

a) The impulse will be computed at the very end of the answer

b) Before the collision

`p_(truck)=2000\cdot 8.2=16400\ kg.m/s`

`p_(car)=1000\cdot 0=0\ kg.m/s`

c) After collision

`p_(truck)=2000\cdot 4=8000\ kg.m/s`

Compute the car's speed:

`\displaystyle v_(2f)=( 16400+0 -8000 )/( 1000 )`

`v_(2f)=8.4\ m/s`

And the car's momentum is

`p_(car)=1000\cdot 8.4=8400\ kg.m/s`

The Impulse J of the system is zero because the total momentum is conserved, i.e. \Delta p=0.

We can compute the impulse for each object

`J_1=\Delta p_1=2000(4-8.2)=-8400 \N.s`

The force can be computed as

`\displaystyle F=(J)/(t)=-(8400)/(0.9)=-9333.33 \ Nw`

The force on the car has the same magnitude and opposite sign