A 799-kg cannon fires a 13.5-kg projectile with a speed of 219 m/s with respect

Question

A 799-kg cannon fires a 13.5-kg projectile with a speed of 219 m/s with respect to the muzzle. The cannon is on wheels and can recoil with negligible friction. Just after the cannon fires the projectile, what is the speed of the projectile with respect to the ground?

Here is my work, I'm not sure where I went wrong

799*v1=-13.5*v2

v1= -0.0169*v2

(using v1+v2=219m/s) I get,

-0.0169*v2+v2= 219

v2=222.76

v1+222.76=219

v1= -3.76 is my answer

I've tried entering that answer in as a positive and negative number. Please help me find the correct answer, thank you

Explanation / Answer

Taking the direction of the projectile as positive
Velocity of projectile = Vp
Velocity of cannon = Vc
Given Relative velocity = Vp w.r.t Vc
= Vp - (-Vc) ( as Vc is opposite to Vp)
=Vp + Vc = 219
==> Vc = 219 - Vp

And applying conservation of linear Momentum
Mp * Vp = - (Mc * Vc)
13.5 * Vp = - (799 * Vc)

Substitute value of Vc = 219 - Vp
13.5 Vp = - 799 * (219 - Vp)

799Vp=13.5 Vp+799*219

785.5 Vp=174981

Vp=222.764

Vp + Vc = 219

Vc=219-222.764

Vc=-3.764

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