Question

A steel ball of 50g initially at rest at point A rolls down an inclined

plane as shown in the figure. What is the speed of the ball when it

reaches the point B? Neglect the frictional force.

(A) 2.38 m/s (B) 5.66 m/s

(C) 8.83 m/s (D) 12.12 m/s​

Answer 1

Explanation:

Force acting on the ball is same in both the cases, therefore acceleration will also be same a

1

=a

2

. Thus, (ii) is wrong.

S=Ut+

2

at

2

Initially ball is at rest, U=0

S=

2

at

2

So, we see that: S∝t

2

Now, as the distance travelled by the two balls are S

1

sinθ=h and S

2

sinθ=2h

S

2

S

1

=

2

1

S

2

=2S

1

So, we get: t

2

=

2

t

1

Thus (i) is wrong.

The acceleration in both cases i.e at position P as well as Q is g, acting in vertically downward direction. Thus (ii) is also wrong.

Now by Conservation of Energy formula, the potential energy possessed by the ball is converted to its Kinetic energy at the bottom of the inclined plane.

PE at point Q = mg(2h) ; PE at point P = mg(h)

Thus, KE at O (when ball falls from Q) = KE at O (when ball falls from P)

So, (iii) is correct.