A block of mass m1 = 19.5 kg slides along a horizontal surface (with friction, μk = 0.35) a distance d = 2.6 m before striking a second block of mass m2 = 8.25 kg. The first block has an initial velocity of v = 6.5 m/s.
(a) Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?
(b) How far does block two travel, d2 in meters, before coming to rest after the collision?

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

[tex]\rm m_1[/tex] is the mass of the block 1 = 19.5 kg

[tex]\rm \mu_k[/tex] is the cofficient of friction =0.35

d is the distance traveled = 2.6 m

m₂ is the mass of block 2 = 8.25 kg.

u is the initial velocity of = 6.5 m/s.

(a)The velocity of block 2 will be 19.5 m/s.

From Newton's second law;

[tex]\rm \sum F= ma \\\\ mg \mu = ma \\\\ a= -g \mu \\\\ a = 9.8 \times 0.22 \\\\\ a=-21.6 m/s^2[/tex]

The velocity of the block before the collision is found with the help of Newton's second equation of motion;

[tex]\rm v^2 = u^2+2as \\\\ \rm v^2 = u^2+2a\triangle x \\\\ v^2 = (8.25)^2+2(-2.16 )(2.3) \\\\ v=7.63\ m/sec[/tex]

The velocity of the block is found with the help of the equation of the law of conservation of momentum;

[tex]\rm m_1u_1 + m_2u_2 = m_2v_2 + m_2v_2 \\\\\ m_1u_1 = m_2v_2(18.5 kg)\times (7.63 m/s) = (7.25 kg) \times v_0 \\\\ v_0 = 19.5 m/s[/tex]

Hence the velocity of block 2 will be 19.5 m/s.

(b)Block 2 will travel a distance of 87.8 m.

From the equation for the second equation of motion;

[tex]v2 = v_0^2+ 2a\triangle x\\\\\ 0 = (19.5)^2 + 2 \times (-2.16 ) \triangle x \\\\ \triangle x = 87.8 m[/tex]

Hence block 2 will travel a distance of 87.8 m.

To learn more about the law of conservation of momentum refer;

https://brainly.com/question/1113396