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A cannon mass 1000 kg located at the the base of an inclined plane fires a shell of mass 100 kg in a horizontal direction with a velocity 180 km/hr. The angle of inclination of inclined plane with horizontal is 45° . The coefficient of friction between the cannon and inclined plane is 0.5. The height in meter to which the cannon ascends the inclined plane as a result of recoil is: (g = 10m/s2)
(1) 7/6
(2) 5/6
(3) 2/6
(4) 1/6

1 Answer
P dilip_k
Jan 19, 2018

drawn

Let

the mass of the cannon be M=1000kg

the mass of the shell be m=100kg

the angle of inclination of the inclined plane be theta=45^@

the coefficient of friction between the cannon and inclined plane be mu=0.5

the velocity of the shell be v=180km"/"hr=(180xx1000)/3600=50m"/"s

If the velocity of recoil of the cannon be V then by conservation of momentum

V=-(mv)/M=-(100xx50)/1000=-5m"/"s
( the negative sign indicates the reverse direction of the cannon w r to the shell.)

If the height of ascent of the center of mass of the cannon over the inclined plane be h m from its initial position, then the slant height covered by the cannon due to its ascent will be L=h/sintheta

By conservation of energy the sum of gravitational potential energy gained by the cannon at height #h m and work done against the frictional force to reach at this height will be equal to the kinetic energy of the cannon due to its velocity of recoil.

So

Mgh+muMgcosthetaxxL=1/2MV^2

=>cancelMgh+mucancelMgcosthetaxxh/sintheta=1/2cancelMV^2

=>gh+mugcotthetaxxh=1/2V^2

=>10h+0.5xx10xxcot45^@xxh=1/2xx5^2

=>15h=25/2

=>h=25/30=5/6 m (OPTION (2)

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