To solve this problem, we need to use the principles of conservation of momentum and energy. The two blocks, A and B, of masses mAm_AmA\u200b and mBm_BmB\u200b, are connected by a spring on a frictionless surface. Initially, the blocks are pulled apart, and the spring is stretched. Upon release, the blocks move, and the spring exerts a force that accelerates both blocks.<\/p>\n\n\n\n
Step 1: Conservation of Momentum<\/h3>\n\n\n\n
Since there are no external forces acting on the system, the total momentum of the system must remain constant. Initially, the blocks are at rest, so the initial momentum is zero. At any instant, let the velocities of blocks A and B be vAv_AvA\u200b and vBv_BvB\u200b, respectively. The conservation of momentum gives the equation:mAvA+mBvB=0m_A v_A + m_B v_B = 0mA\u200bvA\u200b+mB\u200bvB\u200b=0<\/p>\n\n\n\n
This implies:mAvA=\u2212mBvBm_A v_A = -m_B v_BmA\u200bvA\u200b=\u2212mB\u200bvB\u200b<\/p>\n\n\n\n
Thus, the velocities of the blocks are related by:vA=\u2212mBmAvBv_A = -\\frac{m_B}{m_A} v_BvA\u200b=\u2212mA\u200bmB\u200b\u200bvB\u200b<\/p>\n\n\n\n
Step 2: Energy Consideration<\/h3>\n\n\n\n
The total mechanical energy of the system is conserved. The potential energy stored in the spring at any instant is converted into the kinetic energy of the blocks. The total kinetic energy of the system is:K=12mAvA2+12mBvB2K = \\frac{1}{2} m_A v_A^2 + \\frac{1}{2} m_B v_B^2K=21\u200bmA\u200bvA2\u200b+21\u200bmB\u200bvB2\u200b<\/p>\n\n\n\n
Using the relation vA=\u2212mBmAvBv_A = -\\frac{m_B}{m_A} v_BvA\u200b=\u2212mA\u200bmB\u200b\u200bvB\u200b, we substitute vAv_AvA\u200b into the kinetic energy equation:K=12mA(mBmAvB)2+12mBvB2K = \\frac{1}{2} m_A \\left( \\frac{m_B}{m_A} v_B \\right)^2 + \\frac{1}{2} m_B v_B^2K=21\u200bmA\u200b(mA\u200bmB\u200b\u200bvB\u200b)2+21\u200bmB\u200bvB2\u200b<\/p>\n\n\n\n
Simplifying:K=12mA\u22c5mB2mA2vB2+12mBvB2K = \\frac{1}{2} m_A \\cdot \\frac{m_B^2}{m_A^2} v_B^2 + \\frac{1}{2} m_B v_B^2K=21\u200bmA\u200b\u22c5mA2\u200bmB2\u200b\u200bvB2\u200b+21\u200bmB\u200bvB2\u200bK=12(mB2mA+mB)vB2K = \\frac{1}{2} \\left( \\frac{m_B^2}{m_A} + m_B \\right) v_B^2K=21\u200b(mA\u200bmB2\u200b\u200b+mB\u200b)vB2\u200b<\/p>\n\n\n\n
Step 3: Kinetic Energy Ratio<\/h3>\n\n\n\n
The kinetic energy of each block is given by:KA=12mAvA2andKB=12mBvB2K_A = \\frac{1}{2} m_A v_A^2 \\quad \\text{and} \\quad K_B = \\frac{1}{2} m_B v_B^2KA\u200b=21\u200bmA\u200bvA2\u200bandKB\u200b=21\u200bmB\u200bvB2\u200b<\/p>\n\n\n\n
Now, using vA=\u2212mBmAvBv_A = -\\frac{m_B}{m_A} v_BvA\u200b=\u2212mA\u200bmB\u200b\u200bvB\u200b, we can find KAK_AKA\u200b:KA=12mA(mBmAvB)2=12mB2mAvB2K_A = \\frac{1}{2} m_A \\left( \\frac{m_B}{m_A} v_B \\right)^2 = \\frac{1}{2} \\frac{m_B^2}{m_A} v_B^2KA\u200b=21\u200bmA\u200b(mA\u200bmB\u200b\u200bvB\u200b)2=21\u200bmA\u200bmB2\u200b\u200bvB2\u200b<\/p>\n\n\n\n
Thus, the ratio of the kinetic energies is:KAKB=12mB2mAvB212mBvB2=mBmA\\frac{K_A}{K_B} = \\frac{\\frac{1}{2} \\frac{m_B^2}{m_A} v_B^2}{\\frac{1}{2} m_B v_B^2} = \\frac{m_B}{m_A}KB\u200bKA\u200b\u200b=21\u200bmB\u200bvB2\u200b21\u200bmA\u200bmB2\u200b\u200bvB2\u200b\u200b=mA\u200bmB\u200b\u200b<\/p>\n\n\n\n
Therefore, the ratio of the kinetic energies of the two blocks is inversely proportional to the ratio of their masses, as required:KAKB=mBmA\\frac{K_A}{K_B} = \\frac{m_B}{m_A}KB\u200bKA\u200b\u200b=mA\u200bmB\u200b\u200b<\/p>\n\n\n\n
Explanation<\/h3>\n\n\n\n
The key to solving this problem lies in understanding that the system is isolated, so both momentum and energy are conserved. By applying the conservation of momentum, we relate the velocities of the two blocks, and by using energy conservation, we establish the relationship between their kinetic energies. The result shows that the kinetic energy of each block is inversely proportional to its mass, meaning the block with a smaller mass will have a larger kinetic energy.<\/p>\n\n\n\n Two blocks A and B of mass mA and mB are connected together by means of a spring and are resting on a horizontal frictionless table. The blocks are then pulled apart so as to stretch the spring and then released. Show that the ratio of their kinetic energies at any instant is in the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-45076","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45076","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=45076"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45076\/revisions"}],"predecessor-version":[{"id":45086,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45076\/revisions\/45086"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=45076"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=45076"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=45076"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1572.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"