Question

A 2 kg block moving with 10m/s strikes a spring of constant 2n/m attached to 2 kg block of rest kept on a smooth floor. The time for which rear moving block remain in contact with spring will be

Answer 1

Answer:

According to the statement;

m₁ = m₂ = 2 kg

we've a relation for the time period of two spring mass system;

t = (2*π)*√(μ/k) ⇒ (A)

where; k = spring constant = π² N/m

μ = reduced mass = (m₁*m₂)/(m₁+m₂)

the time period of both oscillation same;

(2*π)*√(m₁/k₁) = (2*π)*√(m₂/k₂)

Dividing by 2 on both sides;

(π)*√(m₁/k₁) = (π)*√(m₂/k₂) ⇒ (B)

Now calculations for μ = (m₁*m₂)/(m₁+m₂)

Substituting the values;

μ = (2*2)/(2+2) = 4/4 = 1

μ = 1

Substituting the value of 'μ' & 'k' in (A)

t = (2*π)*√(1/π²)

t = 2 seconds

But (B) implies;

Time = 0.5*2 = 1 second

The time for which rear moving block remain in contact with spring will be 1 second