A block of mass m 1 =1.6kg initially moving to the right with a speed of3.4m/s o
ID: 2992737 • Letter: A
Question
A block of massm1=1.6kg initially moving to the right with a speed of3.4m/s on a frictionless, horizontal track collides with a spring attached to a second block of massm2=3.2kg initially moving to the left with a speed of2.4m/s as shown in figure (a). The spring constant is552N/m.
(A) Find the velocities of the two blocks after the collision.
(B) During the collision, at the instant block 1 is moving to the right with a velocity of1.0m/s as in figure (b), determine the velocity of block 2.
(C) Determine the distance the spring is compressed at that instant.
What ifm1is initially moving at3.2m/s whilem2is initially at rest?
A block of massm1=1.6kg initially moving to the right with a speed of3.4m/s on a frictionless, horizontal track collides with a spring attached to a second block of massm2=3.2kg initially moving to the left with a speed of2.4m/s as shown in figure (a). The spring constant is552N/m. (A) Find the velocities of the two blocks after the collision. (B) During the collision, at the instant block 1 is moving to the right with a velocity of1.0m/s as in figure (b), determine the velocity of block 2. (C) Determine the distance the spring is compressed at that instant. What ifm1is initially moving at3.2m/s whilem2is initially at rest? (a) Find the maximum spring compression in this case. x= m (b) What will be the individual velocities of the two masses (v1andv2) after the spring extended fully again? (That is, when the two masses separate from each other after the collision is complete.) v1=m/s to the left v2=m/s to the rightExplanation / Answer
b) First, note that the initial velocity of block 2 is -2.40 m/s because its direction is to the left. Because momentum is conserved for the system of two blocks, we have m1v1i+ m2v2i = m1v1f + m2v2f (1.60 kg)(3.4 m/s) + (3.20 kg)(-2.4 m/s)= (1.60 kg)(1.00 m/s)+(3.20 kg)v2f v2f = - 1.2 m/s The negative value for v2f means that block 2 is still moving to the left at the instant we are considering. c) To determine the distance that the spring is compressed, shown as x in Figure b, we can use the con-cept of conservation of mechanical energy because no friction or other nonconservative forces are acting on the system. Thus, we have 1/2 m1v1i ^2 + 1/2 m2v2i ^2 = 1/2 m1v1f ^2+1/2 m2v2f ^2 +1/22 kx^2 Substituting the given values and the result to part (b) into this expression gives (0.5)(1.60 kg)(3.4 m/s)^2 + 0.5* (3.2 kg)(-2.4 m/s)^2= 0.5*(1.60 kg)(3.2 m/s)^2+0.5* (3.2 kg)(-1.2 m/s)^2+0.5*552*x^2 x =0.0746 m It is important to note that we needed to use the principles of both conservation of momentum and conservation of mechanical energy to solve the two parts of this problem.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.