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Butoxors [25]
3 years ago
6

A 2,000 kg car is moving at 15m/s when it collides with a 1200kg car sitting still. Briefly compare the impulse imparted on the

1200kg car by the 2000 kg car to the impulse on the 2000 kg car by the 1200 kg car which car under goes the greater change in momentum
Physics
2 answers:
mihalych1998 [28]3 years ago
8 0

Answer:

Explanation:

Given:

mass of car 1 = 2000 kg

speed of car 1 = 15 m/s

mass of car 2 = 1200 kg

car 2 is initially at rest.

Impulse (F Δt) = change in momentum = Δ(mv)

Change in Momentum depends on the mass and velocity.

When car 1 collides with car 2, car 1 would under go greater change in the momentum because it has greater mass and initial velocity while car 2 has smaller mass and is initially at rest.

I am Lyosha [343]3 years ago
4 0

impulse = F × t

The greater the impulse exerted on something, the greater will be the change in momentum.

impulse = change in momentum

Ft = ∆(mv)

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Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is
Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s​2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is  c = 28853.78 \ m^2 /s^2

Explanation:

From the question we are told that

         The acceleration is  a =  \frac{c}{s}\   m/s^2

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is s_f =  3 \ m

          The velocity is  v = 200 \ m/s

     Generally  time  =  \frac{ds}{dv}

   and  acceleration is a =  \frac{v}{time }

so

            a = v  \frac{dv}{ds}

 =>        vdv  =  a ds

             vdv  = \frac{c}{s}  ds

integrating both sides

           \int\limits^a_b  vdv  = \int\limits^c_d \frac{c}{s}  ds

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =s_f= 1.5 m

So we have  

           \int\limits^{200}_{0}  vdv  = \int\limits^{3}_{1.5} \frac{c}{s}  ds

              [\frac{v^2}{2} ] \left | 200} \atop {0}} \right.  = c [ln s]\left | 3} \atop {1.5}} \right.

            \frac{200^2}{2}  =  c ln[\frac{3}{1.5} ]

=>           c = \frac{20000}{0.69315}

              c = 28853.78 \ m^2 /s^2

     

5 0
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