Question

A 15-lb block B starts from rest and slides on the 25-lb wedge A, which is supported by a horizontal surface. Neglecting friction, determine (a) the velocity of B relative to A after it has slid 3 ft down the inclined surface of the wedge, (b) the corresponding velocity of A.

Answer 1

The velocity of block B relative to wedge A is approximately 10 ft/s. The velocity of wedge A is 0 ft/s.

Step-by-step explanation:

To solve this problem, we can use the principles of conservation of momentum and conservation of energy. Since the wedge A is supported by a horizontal surface and there is no friction, the only force acting on the system is the gravitational force. We can calculate the velocity of block B relative to wedge A using the principle of conservation of energy. The gravitational potential energy of block B at the starting position is converted to kinetic energy as it slides down the inclined surface of the wedge. The equation for the conservation of energy is:

KE_start = PE_start

½mv_start^2 = mgh_start

½(15 lb)(v_AB)^2 = (15 lb)(32.2 ft/s^2)(3 ft)

(v_AB)^2 = 2(32.2)(3)

v_AB = √(2(32.2)(3))

v_AB ≈ 10 ft/s

To find the velocity of wedge A, we can use the principle of conservation of momentum. Since there is no external force acting on the system, the total momentum before and after the motion remains constant. The total momentum of the system before the motion is zero because both block B and wedge A are at rest. After the motion, the two objects move together as one, so their velocities are equal. We can set up the equation:

0 = (15 lb)(v_A)

v_A = 0

Answer 2

The angle of the wedge is 30°.

Answer:

5.88 ft/s

Step-by-step explanation:

a) The block will slide down due to it's weight.

initial velocity u= 0

final velocity, v

acceleration, a = g sin 30° = 32 ft/s²× sin 30° = 16 ft/s²

Sliding displacement, s = 3ft

Use third equation of motion:

`v^2-u^2 = 2as`

substitute the values and solve for v

`v^2-0 = 2* 16 * 3 =96 ft^2/s^2\\v = 9.8 ft/s`

b) Use conservation of momentum:

Initial momentum of the system = 0

final momentum = (15) ( 9.8)+ (25)(v')

v' = 5.88 ft/s