A 1 kg ball is fired from a cannon inclined at 45 degrees relative to the horizo

Question

A 1 kg ball is fired from a cannon inclined at 45 degrees relative to the horizon. The firing mechanism is a spring initially-compressed by 0.5 m; the strength constant of the spring is 10, 000 N/m; the coefficient of kinetic friction between the ball and the barrel of the cannon is 0.2; and the initial distance from the ball to the end of the barrel is 8 m.

Please explain in detail to get all points.

a. What is the velocity of the call as it exits the barrel?

b. If the cannon has a mass of 20 kg and sits on a frictionless platform, what is its velocity as the ball exits its barrel?

Explanation / Answer

m = 1 kg

x = 0.5m

k = 10,000 N/m

angle = 45 degree

uk = 0.2

by energy conservation

1/2 *k*x^2 = uk *N*8 + 1/2*m*v^2

0.5*10000 *0.5^2 = 0.2 *mg*cos45 *8 + 0.5 *1 *v^2

v = 49.77 m/s

*********************

by momentum conservation

Mcannon*V = m*v

20 *Vcannon = 1 *49.77

V cannon = 2.49 m/s

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