Question #193654

7.Police suspect that a van was breaking the speed limit when it collided with a parked car. Investigations provided them with the following data: mass of parked car                  = 800 kg,  mass of van = 1000 kg,  velocity of  car after collision = 5 m/s,  velocity of van after collision = 11 m/s


Show whether the van was breaking the 30 mph speed limit (equivalent to 13 m/s).


8.A dodgem car of mass 200 kg moves with a velocity of 3 m/s. The moving dodgem collides with a stationary one of mass 180 kg.


After the collision, the 200 kg dodgem continues to move in the same direction at a speed of 1 m/s.


Calculate the speed v with which the 180 kg (second) dodgem moves to the right after the collision.





1
Expert's answer
2021-05-16T17:59:04-0400

7.7.

van: \text{van: }

m1=1000;u1?;v1=11m_1 =1000 ;\vec u_1-?;\vec v_1= 11

car:\text{car:}

m2=800;u2=0;v2=5m_2 =800;\vec u_2=0;\vec v_2= 5

u1,u2 pre-collision speed\vec u_1,\vec u_2\ \text{pre-collision speed}

v1,v2 speed after collision\vec v_1, \vec v_2\ \text{speed after collision}

according to the law of conservation of momentum:\text{according to the law of conservation of momentum:}

p=m1u1+m2u2 =m1u1=1000u1\vec p= m_1\vec u_1+m_2\vec u_2\ = m_1\vec u_1=1000\vec u_1

p=m1v1+m2v2 =100011+8005=15000\vec p= m_1\vec v_1+m_2\vec v_2\ = 1000*11+800*5=15000

1000u1=150001000\vec u_1= 15000

u1=15ms>13ms\vec u_1= 15\frac{m}{s}>13\frac{m}{s}

Answer: The van driver has exceeded the speed limit\text{Answer: The van driver has exceeded the speed limit}


8.8.

dodgem car first:\text{dodgem car first:}

m1=200;u1=3;v1=1m_1 =200 ;\vec u_1=3;\vec v_1= 1

dodgem car second:\text{dodgem car second:}

m2=180;u2=0;v2?m_2 =180 ;\vec u_2=0;\vec v_2-?

u1,u2 pre-collision speed\vec u_1,\vec u_2\ \text{pre-collision speed}

v1,v2 speed after collision\vec v_1, \vec v_2\ \text{speed after collision}

according to the law of conservation of momentum:\text{according to the law of conservation of momentum:}

p=m1u1+m2u2 =2003=600\vec p= m_1\vec u_1+m_2\vec u_2\ = 200*3=600

p=m1v1+m2v2=2001+180v2\vec p= m_1\vec v_1+m_2\vec v_2= 200*1+180\vec v_2

2001+180v2=600200*1+180\vec v_2 =600

v2=4001802.22ms\vec v_2=\frac{400}{180}\approx2.22\frac{m}{s}

Answer: 2.22ms\text{Answer: }\approx2.22\frac{m}{s}


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