Question

1- A ball with mass M, moving horizontally at 2.8 m/s, collides elastically with a block with mass 3.6M that is initially hanging at rest from the ceiling on the end of a 58-cm wire. Find the maximum angle through which the block swings after it is hit, in degrees.

2- A 0.15 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.72 m/s. It has a head-on collision with a 0.30 kg glider that is moving to the left with a speed of 0.36m/s. Suppose the collision is elastic. Find the magnitude of the final velocity (in m/s) of the 0.150kg glider.

Answer 1

1.

m_b = mass of the ball = M

m_B = mass of the block = 3.6 M

v_bo = initial velocity of ball before collision = 2.8 m/s

v_b = velocity of ball after collision

V_B = velocity of the block after collision

Using conservation of momentum

m_b v_bo = m_b v_b + m_B V_B

M (2.8) = M v_b + (3.6) M V_B

2.8 = v_b + (3.6) V_B

v_b = 2.8 - (3.6) V_B Eq-1

using conservation of kinetic energy

m_b v²_bo = m_b v²_b + m_B V²_B

M (2.8)² = M v²_b + (3.6) M V²_B

(2.8)² = v²_b + (3.6) V²_B

Using eq-1

(2.8)² = (2.8 - (3.6) V_B )² + (3.6) V²_B

V_B = 1.22 m/s

h = height gained by the block

Using conservation of energy for the block after the collision

Kinetic energy at the bottom just after collision = potential energy gained by the block

(0.5) m_B V_B² = m_B g h

h = (0.5) V_B² /g

h = (0.5) (1.22)² /9.8

h = 0.076 m

= angle through which the block swings

L = length of the wire = 58 cm = 0.58 m

height gained is also given as

h = L (1 - cos)

0.076 = 0.58 (1 - cos)

cos = 0.87

= Cos^-1(0.87)

= 29.5 deg