A steel ball of mass, m, is fastened to a cord that is L in length and fixed at the far end. The ball is then released when the cord is horizontal. At the bottom of its path, the ball strikes a steel block with a mass of M. Find: (a) the speed of the ball (b) the speed of the black after the collision. (c): If the block flies off the table, what is the distance it traveled in the positive x-direction?
The correct answer and explanation is :
To solve this problem, we’ll go step by step. Let’s define the knowns and apply physics principles.
Given:
- Mass of ball = ( m )
- Cord length = ( L )
- Mass of block = ( M )
- The ball is released from rest at the height ( h = L )
- Collision occurs at the lowest point (bottom of swing)
- The collision is perfectly elastic (since it’s between two steel objects)
- The block is on a frictionless table and can fly off the edge
(a) Speed of the ball at the bottom of its path
Use conservation of mechanical energy:
[
\text{Initial Potential Energy} = \text{Kinetic Energy at bottom}
]
[
m g L = \frac{1}{2} m v^2
\Rightarrow v = \sqrt{2gL}
]
(b) Speed of the block after the collision
Assuming a perfectly elastic collision and linear momentum conservation:
Let:
- ( u_1 = \sqrt{2gL} ) = initial speed of the ball
- ( u_2 = 0 ) = block initially at rest
- ( v_1 ), ( v_2 ) = final velocities of ball and block respectively
Apply conservation of momentum and kinetic energy:
Momentum:
[
mu_1 + Mu_2 = mv_1 + Mv_2
\Rightarrow m\sqrt{2gL} = mv_1 + Mv_2 \quad (1)
]
Kinetic energy:
[
\frac{1}{2}m u_1^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}Mv_2^2 \quad (2)
]
Solving these equations gives:
[
v_2 = \frac{2m}{m+M} \sqrt{2gL}
]
(c) Distance traveled by the block in x-direction
After the collision, the block leaves the edge of the table horizontally with speed ( v_2 ). Let table height be ( H ). The time to fall:
[
t = \sqrt{\frac{2H}{g}}
]
Horizontal distance traveled:
[
x = v_2 \cdot t = \frac{2m}{m+M} \sqrt{2gL} \cdot \sqrt{\frac{2H}{g}} = \frac{2m}{m+M} \sqrt{4LH}
]
[
x = \frac{4m}{m+M} \sqrt{LH}
]
Summary:
- (a) Ball speed at bottom: ( \boxed{v = \sqrt{2gL}} )
- (b) Block speed after collision: ( \boxed{v_2 = \frac{2m}{m+M} \sqrt{2gL}} )
- (c) Horizontal distance: ( \boxed{x = \frac{4m}{m+M} \sqrt{LH}} )
Explanation (300 words):
This problem explores energy conservation, momentum, and projectile motion in a classic physics setup. Initially, the steel ball is raised to a height ( L ), meaning it has gravitational potential energy ( m g L ). When released, this potential energy converts entirely into kinetic energy at the lowest point of the swing because there’s no friction or air resistance. Using energy conservation, we determine that the ball’s speed at this point is ( \sqrt{2gL} ).
Next, we examine the collision. Because it’s between steel objects and we assume no energy loss, the collision is elastic. In such collisions, both momentum and kinetic energy are conserved. Using these two principles, we solve for the final velocities of both objects. The resulting speed of the block after collision is determined using algebraic manipulation of the conservation equations, and turns out to be ( \frac{2m}{m+M} \sqrt{2gL} ).
Finally, when the block flies off the edge of the table, it undergoes projectile motion. It moves horizontally with constant velocity (no air resistance), while simultaneously falling under gravity. The time it takes to hit the ground depends only on the height ( H ) of the table. Multiplying this fall time by the block’s horizontal speed gives the distance traveled in the x-direction. The final result, ( \frac{4m}{m+M} \sqrt{LH} ), shows how both masses and table height affect how far the block flies.
This problem beautifully integrates multiple core physics concepts in one scenario.
