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myrzilka [38]
3 years ago
9

Issac and Blaise decide to race. They both start at the same position at the same time. Issac runs at 2m/s but decides to take a

2 minute rest stop along the way. Blaise runs at half the speed and still wins by 10 m. How far did Blaise run? Please show your work, it’ll greatly help with my understanding. I know I have to do some substitution with equations, but I don’t have a final length of the race, or final time, so I have no idea where to start.
Physics
1 answer:
FromTheMoon [43]3 years ago
6 0

Let the Blaise runs for time "t" to complete the race

so the total distance he moved is given by

d_1 = 1* t

Now Issac runs for time t = "t - 2*60"

because it took rest for 2 minutes

d_2 = 2*(t - 120)

now it is given that Blaise wins by 10 m distance

d_1 - d_2 = 10

1* t - 2*(t - 120) = 10

t - 2t + 240 = 10

t = 230 s

now the distance moved by Blaise is given by

d_1 = 1*230 = 230 m

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Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

U=(M+m)gh       (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m

The potential energy is:

U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

Mv_1+mv_2=(M+m)v    (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

M(0)+mv_2=(M+m)v    

v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N    

The force on the cord after the impact is 2.59N

4 0
3 years ago
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