B.1 () A solid sphere of mass 2.2 kg and radius 4.5 cm rolls without slipping do

Question

B.1 () A solid sphere of mass 2.2 kg and radius 4.5 cm rolls without slipping down a ramp. At the bottom of the ramp, the sphere encounters a small lip inclined at an angle of 18° with respect to the horizontal, which launches the sphere into the air. If the ramp has a height of 120 cm, calculate how far away from the end of the ramp the sphere will land. The moment of inertia for a solid sphere is given by 1-Mr2 (ii) A bullet of mass 115 g is fired at a wooden block of mass 2.6 kg and becomes embedded in the block. The wooden block is resting on a rough surface and is attached to an uncompressed spring which is mounted on a wall. The spring is a Hookean spring with a spring constant k = 540 N/m and the coefficient of friction between the block and the surface is -0.35. If the spring compresses a maximum distance of 33 cm after the bullet is fired into the block, what was the initial velocity of the bullet?

Explanation / Answer

part i.

as there is no friction, total energy will be conserved.

let speed at the bottom is v m/s.

at the top of the ramp, velocity is zero hence total energy=potential energy=mass*g*height

at the bottom of the ramp, height is zero hence total energy=linear kinetic energy + rotational kinetic energy

=0.5*mass*speed^2+0.5*moment of inertia*angular speed^2

=0.5*mass*v^2+0.5*(0.4*mass*radius^2)*(v/radius)^2

=0.7*mass*v^2

equating both the energy values:

mass*g*height=0.7*mass*v^2

==>v=sqrt(g*height/0.7)=sqrt(9.8*1.2/0.7)=4.0988 m/s

launch angle=18 degree

then range of this projectile=initial speed^2*sin(2*launch angle)/g

=4.0988^2*sin(2*18)/9.8

=1.0076 m

so the sphere will land at a distance of 1.0076 m.

part b:

mass of bullet=m=0.115 kg

mass of block=M=2.6 kg

spring constant=k=540 N/m

friction coefficient=mu=0.35

spring compression=x=0.33 m

let initial speed of the bullet be v.

normal force on block -bullet system=mass*g=(M+m)*g

friction force=friction coefficient*normal force=mu*(M+m)*g

using work energy principle:

initial kinetic energy of the bullet-work done against friction=final potential energy of the spring

=>0.5*m*v^2-force*distance=0.5*spring constant*compression^2

==>0.5*m*v^2-mu*(m+M)*g*x=0.5*k*x^2

==>v=sqrt((0.5*k*x^2+mu*(M+m)*g*x)/(0.5*m))=23.766 m/s

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