Question

A 1.5kg block slides along a frictionless surface at 1.3m/s . A second block, sliding at a faster 4.3m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0m/s . What was the mass of the second block?

Answer 1

The mass of the second block is approximately 0.457 kg.

Step-by-step explanation:

In this problem, we can use the principle of conservation of momentum to find the mass of the second block. Before the collision, the total momentum of the system is given by the sum of the individual momenta of the two blocks:

1.5 kg * 1.3 m/s + m2 * 4.3 m/s = (1.5 kg + m2) * 2.0 m/s

Simplifying this equation, we can solve for m2:

1.95 kg*m/s + 4.3 m2 = 3 kg*m/s + 2m2

Subtracting 2m2 from both sides, we have:

1.95 kg*m/s + 2.3 m2 = 3 kg*m/s

Subtracting 1.95 kg*m/s from both sides, we get:

2.3 m2 = 1.05 kg*m/s

Finally, dividing both sides by 2.3, we find that:

m2 ≈ 0.457 kg

Answer 2

The mass of the second block=0.457 kg

Step-by-step explanation:

We are given that

m1=1.5 kg

v1=1.3m/s

v2=4.3 m/s

V=2.0 m/s

We have to find the mass of the second block.

`m_1v_1+m_2v_2=(m_1+m_2)V`

Let m2=m

Substitute the values

`1.5(1.3)+m(4.3)=(1.5+m)(2)`

`1.95+4.3m=3+2m`

`4.3m-2m=3-1.95`

`2.3m=1.05`

`m=(1.05)/(2.3)`

`m=0.457 kg`

Hence, the mass of the second block=0.457 kg