Question

Three identical blocks A, B and C are placed on horizontal frictionless surface. The blocks A and C are at rest. But A is approaching towards B with a speed 10 m/s. The coefficient of restitution for all collision is 0.5. The speed of the block C just after the collision is ______.

Options

• 5.6 m/s

• 6 m/s

• 8 m/s

• 10 m/s

Answer 1

Three identical blocks A, B and C are placed on horizontal frictionless surface. The blocks A and C are at rest. But A is approaching towards B with a speed 10 m/s. The coefficient of restitution for all collision is 0.5. The speed of the block C just after the collision is 5.6 m/s.

Explanation:

For collision between block A and B,

e = `("v"_"B"-"v"_"A")/("u"_"A"-"u"_"B")=("v"_"B"-"v"_"A")/(10-0)=("v"_"B"-"v"_"A")/10`

∴ v_B - v_A = 10e = 10 × 0.5 = 5       ...(i) 

From principle of momentum conservation,

m_Au_A + m_Bu_B = m_Av_A + m_Bv_B 

or m × 10 + 0 = mv_A + mv_B

∴ v_A + v_B = 10                      ...(ii)

Adding Eqs. (i) and (ii), we get

v_B = 7.5 ms^-1                        ...(iii)

Similarly for collision between B and C,

v_C - v_B ​= 7.5e = 7.5 × 0.5 = 3.75

∴​ v_C - v_B ​= 3.75 ms^-1                ...(iv)

Adding Eqs. (iii) and (iv), we get

2v_C = 11.25

v_C = `11.25/2`

= 5.6 ms^-1