Calculate the final velocity right after a 117 kg rugby player who is initially running at 7.45 m/s collides head‑on with a padded goalpost and experiences a backward force of 17700 N for 5.50×10−2 s

The final velocity, right after a 117 kg rugby player who is initially running at 7.45 m/s collides head‑on with a padded goalpost when it experiences a backward force of 17700 N for 5.50×10−2 s is 1.95 m/s

Applying the formula of impulse,

Ft = m(v-u) Equation 1

Where F = force, t = time, m = mass, v = final velocity, u = initial velocity.

From the question,

GIven: F = -17700 N ( negative because it's backward force), t = 5.5×10⁻² s, m = 117 kg, u = 7.45 m/s,

Substitute these values into equation 1

-17700(0.055) = 177(v-7.45)

v-7.45 = -(17700×0.055)/177

v-7.45 = -5.5

v = 7.45-5.5

v = 1.95 m/s.

Hence, the final velocity of the rugby player is 1.95 m/s

Learn more about impulse of a force here: link

v_f = 0.87 m/s

Explanation:

We are given;

F_avg = -17700 N (negative because it's backward)

m = 117 kg

Δt = 5.50 × 10^(−2) s

v_i = 7.45 m/s

Now, formula for impulse is given by;

I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s

From impulse momentum theory, we know that;

Change in momentum of particle is equal to impulse.

Thus,

Δp = I = m•v_f - m•v_i

Thus,

-973.5= 117(v_f - 7.45)

Thus,

-973.5/117 = (v_f - 7.45)

-8.3205 + 7.45 = v_f

v_f = - 0.87 m/s

We'll take absolute value as;

v_f = 0.87 m/s

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j