As in procedure 1: a pair of carts sit on a frictionless track. Assume the mass

Question

As in procedure 1: a pair of carts sit on a frictionless track. Assume the mass of the larger cart is5.17times the mass of the small cart, In this system, however, there is a spring between the carts which can "explode", pushing on the two carts at once..

NOTE: Every velocity needs magnitude and direction (given by the sign).

a) Suppose the carts are initially at rest, and after the "explosion" the smaller cart is moving at velocity +3.93m/s.

- Find the velocity of the larger cart.______m/s

- If the mass of the smaller cart is5.45kg, find the energy supplied by the spring to the carts._______J

- If the spring has spring constant k =842N/m, find the distance the spring was compressed before the "explosion"._______m

HINT: Think conservation of energy!


b) Suppose the carts are initially moving together at constant velocity +8.14m/s. After the "explosion", the smaller cart is moving at velocity +3.93m/s. Find the velocity of the larger cart.
________m/s

Explanation / Answer

A pair of carts sit on a frictionless track. Assume the mass of the larger cart is 3.48 times the mass of the? small cart, In this system, however, there is a spring between the carts which can "explode", pushing on the two carts at once.. NOTE: Every velocity needs magnitude and direction (given by the sign). let mass of small cart = m mass of larger cart = 5.17 m a) Suppose the carts are initially at rest, and after the "explosion" the smaller cart is moving at velocity +3.93 m/s. Momentum is always conserved! Initial momentum = 0, since both carts are at rest Final momentum = (m * 3.93) + (5.17 * m * v) Final momentum = Initial momentum (m * 3.93) + (5.17 * m * v) = 0 (m * 3.93) =- (5.17 * m * v) divide both sides by m 3.93 = -(5.17 * v) v = -0.760 m/s The velocity of the larger cart = 0.760 m/s in the opposite direction of smaller car! - If the mass of the smaller cart is 5.45 kg, find the energy supplied by the spring to the carts. J The energy supplied by the spring is transferred to the carts. Total KE of carts = energy supplied by the spring KE = ½ * mass * velocity^2 Mass of larger car = 5.17 * 5.45 = 28.176 kg Total KE = ½ * 5.45 * 3.93^2 + ½ * 28.176 *0.760^2 The energy supplied by the spring =½ * 5.45 * 3.93^2 + ½ * 28.176 *0.760^2 =50.22 Spring potential energy = ½ * k * distance^2 ½ * 842 * distance^2 = 50.22 distance = (2 * 50.22 ÷ 842)=.345 - If the spring has spring constant k = 842 N/m, find the distance the spring was compressed before the "explosion". m HINT: Think conservation of energy! b) Suppose the carts are initially moving together at constant velocity +8.14 m/s. After the "explosion", the smaller cart is moving at velocity +3.93 m/s. Find the velocity of the larger cart. m/s Total Initial momentum = (5.45+ 28.176) * 8.14 Total Final momentum = (5.45 * 3.93) + (28.176 * vf) (5.45 * 3.93) + (28.176 * vf) = (5.45+ 28.176) * 8.14 solve for vf vf =8.925m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.