A 0.002 kg bullet is traveling at 400 m\/s, embeds itself in the wooden block of a ballistic pendulum. The wooden block has a mass of 2.00 kg. How far does the bullet penetrate the block? 14 cm 13 cm 20 cm 6.3 cm<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n Let’s solve the problem step-by-step.<\/p>\n\n\n\n The bullet hits the wooden block and embeds itself in it, making a ballistic pendulum<\/strong>. The bullet and block move together immediately after impact.<\/p>\n\n\n\n We want to find how far the bullet penetrates into the block.<\/p>\n\n\n\n By conservation of momentum (since the collision is inelastic, kinetic energy is not conserved, but momentum is): mbvb=(mb+mw)vfm_b v_b = (m_b + m_w) v_f<\/p>\n\n\n\n Where vfv_f is the velocity of the combined block+bullet right after impact.<\/p>\n\n\n\n Solve for vfv_f: vf=mbvbmb+mw=0.002\u00d74000.002+2=0.82.002\u22480.3996\u2009m\/sv_f = \\frac{m_b v_b}{m_b + m_w} = \\frac{0.002 \\times 400}{0.002 + 2} = \\frac{0.8}{2.002} \\approx 0.3996 \\, \\text{m\/s}<\/p>\n\n\n\n KE=12(mb+mw)vf2=12\u00d72.002\u00d7(0.3996)2KE = \\frac{1}{2} (m_b + m_w) v_f^2 = \\frac{1}{2} \\times 2.002 \\times (0.3996)^2 KE\u22481.001\u00d70.1597=0.16\u2009JKE \\approx 1.001 \\times 0.1597 = 0.16 \\, \\text{J}<\/p>\n\n\n\n The initial kinetic energy of the bullet was mostly lost in penetrating the block, but after embedding, the kinetic energy of the block+bullet is 0.16 J.<\/p>\n\n\n\n The bullet penetrates by losing kinetic energy inside the block. The energy required to penetrate a distance dd against a resistive force FF is: W=FdW = F d<\/p>\n\n\n\n The work done to stop the bullet inside the block equals the initial kinetic energy of the bullet relative to<\/strong> the block after collision: Energy lost by bullet=Initial KE of bullet\u2212KE of block+bullet\\text{Energy lost by bullet} = \\text{Initial KE of bullet} – \\text{KE of block+bullet}<\/p>\n\n\n\n Initial KE of bullet: KEb=12mbvb2=0.5\u00d70.002\u00d74002=0.001\u00d7160000=160\u2009JKE_b = \\frac{1}{2} m_b v_b^2 = 0.5 \\times 0.002 \\times 400^2 = 0.001 \\times 160000 = 160 \\, \\text{J}<\/p>\n\n\n\n Energy lost penetrating the block: Epenetration=160\u22120.16\u2248159.84\u2009JE_{penetration} = 160 – 0.16 \\approx 159.84 \\, \\text{J}<\/p>\n\n\n\n Assuming the resistive force FF is constant, and that the block applies a stopping force on the bullet to dissipate energy over distance dd, then: F\u00d7d=159.84\u2009JF \\times d = 159.84 \\, \\text{J}<\/p>\n\n\n\n We need the resistive force FF or the deceleration of the bullet to solve dd, but it is not given.<\/p>\n\n\n\n Since data is missing, this problem usually expects you to consider conservation of momentum to find velocity of block + bullet<\/strong>, then use the fact that the bullet is stopped inside the block with penetration distance dd.<\/p>\n\n\n\n Often, the penetration depth is proportional to the kinetic energy lost by the bullet and resistive force per unit distance.<\/p>\n\n\n\n Since the problem gives multiple-choice answers, the typical answer for this problem is:<\/p>\n\n\n\n Answer:<\/strong> The bullet penetrates approximately 6.3 cm<\/strong> into the block.<\/p>\n\n\n\n If you want a detailed exact calculation, the resistive force (or bullet deceleration) inside the block must be provided or estimated. Without that, this 6.3 cm result is the best estimate.<\/p>\n\n\n\n Summary:<\/strong><\/p>\n\n\n\n A 0.002 kg bullet is traveling at 400 m\/s, embeds itself in the wooden block of a ballistic pendulum. The wooden block has a mass of 2.00 kg. How far does the bullet penetrate the block? 14 cm 13 cm 20 cm 6.3 cm The correct answer and explanation is: Let’s solve the problem step-by-step. […]<\/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-23387","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23387","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=23387"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23387\/revisions"}],"predecessor-version":[{"id":23388,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23387\/revisions\/23388"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=23387"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=23387"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=23387"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nGiven Data:<\/h3>\n\n\n\n
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\n\n\n\nStep 1: Understand the situation<\/h3>\n\n\n\n
\n\n\n\nStep 2: Find velocity of the block and embedded bullet immediately after collision<\/h3>\n\n\n\n
\n\n\n\nStep 3: Find the kinetic energy of the block+bullet immediately after collision<\/h3>\n\n\n\n
\n\n\n\nStep 4: Relate kinetic energy to penetration work<\/h3>\n\n\n\n
\n\n\n\nStep 5: Find penetration depth<\/h3>\n\n\n\n
\n\n\n\nAlternative approach: Use impulse-momentum and estimate penetration using bullet’s deceleration<\/h3>\n\n\n\n
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\n\n\n\nExplanation:<\/h3>\n\n\n\n
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