Question

3.10.11 un pendule balistique. le pendule balistique représenté sur le schéma ci- comme sert à déterminer expérimen- talement le module de la vitesse v d’une balle de fusil. la balle de masse ( m_1 ) pénètre et s’incruste dans un bloc de bois de masse ( m_2 ) accroché à deux cordes de longueur ( l ) (on utilise deux cordes pour empêcher le bloc de tourner sur lui-même après l’impact). une petite cale de masse négligeable est placée derrière le bloc et se déplace d’une distance ( x ) lorsque le pendule atteint sa hauteur maximale. si ( x = 20 , \text{cm} ), ( l = 80 , \text{cm} ), ( m_1 = 0,01 , \text{kg} ) et ( m_2 = 5 , \text{kg} ), déterminez ( v ). 3.10.12 deux rondelles rebondissent l’une sur vue du dessus

Explanation:

Step1: Find the height \( h \) the pendulum rises

We can use the Pythagorean theorem. The length of the cord is \( L = 80\space cm=0.8\space m \), and the horizontal displacement is \( x = 20\space cm = 0.2\space m \). Let the height risen be \( h \). Then, from \( L^{2}=(L - h)^{2}+x^{2} \), expanding we get \( L^{2}=L^{2}- 2Lh+h^{2}+x^{2} \). Since \( h \) is small, \( h^{2}\) is negligible, so \( 0=-2Lh + x^{2}\), then \( h=\frac{x^{2}}{2L} \). Substituting \( x = 0.2\space m \) and \( L=0.8\space m \), we have \( h=\frac{(0.2)^{2}}{2\times0.8}=\frac{0.04}{1.6} = 0.025\space m \).

Step2: Apply conservation of mechanical energy for the pendulum

After the impact, the combined mass \( M=m_{1} + m_{2}\) moves and rises to height \( h \). Using conservation of mechanical energy, the kinetic energy after impact is converted to gravitational potential energy. So \( \frac{1}{2}(m_{1}+m_{2})v_{f}^{2}=(m_{1} + m_{2})gh \), where \( v_{f}\) is the velocity of the combined mass just after impact. We can cancel \( (m_{1}+m_{2}) \) from both sides, so \( v_{f}=\sqrt{2gh} \). Substituting \( g = 9.8\space m/s^{2}\) and \( h = 0.025\space m \), we get \( v_{f}=\sqrt{2\times9.8\times0.025}=\sqrt{0.49}=0.7\space m/s \).

Step3: Apply conservation of momentum for the in - elastic collision

Before the collision, the momentum of the system is \( p_{i}=m_{1}v \) (since the block of mass \( m_{2} \) is at rest initially). After the collision, the momentum of the system is \( p_{f}=(m_{1}+m_{2})v_{f} \). By conservation of momentum \( m_{1}v=(m_{1}+m_{2})v_{f} \). We know \( m_{1}=0.01\space kg \), \( m_{2}=5\space kg \), and \( v_{f} = 0.7\space m/s \). Solving for \( v \), we have \( v=\frac{(m_{1}+m_{2})v_{f}}{m_{1}} \). Substituting the values: \( m_{1}+m_{2}=0.01 + 5=5.01\space kg \), so \( v=\frac{5.01\times0.7}{0.01}=\frac{3.507}{0.01}=350.7\space m/s \approx351\space m/s \)

Answer:

The velocity \( v \) of the bullet is approximately \(\boldsymbol{351\space m/s}\) (or more precisely \( 350.7\space m/s \))